You must have 100 questions here. :-).
We don't like to answer so many on one post. It takes too much time.
For #1 I don't agree with your answers. Booth appear to be off by a factor of 10.
5/366.73 = mols Sn(NO3)4 = ?
M = ?/0.250 = about 0.0545M
Then 0.0545M x (15/90) = 0.00909M Sn(NO3)4.
Then Sn^4+ = 0.00909M
NO^ = 4 x 0.00909 = 0.0364
******* Please check. Also check sig figs.
Ion Concentrations
1.) A solution is prepared by dissolving 5.00 g of stannic nitrate in enough water to make 250.0 mL of stock solution. A 15.0 mL aliquot (portion) of this stock solution is then removed and added to 75.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[Sn(NO3)4]=0.00909M
[Sn4+]= 0.000909M
[NO3-]= 0.00364M
2.) A solution is prepared by dissolving 5.00 g of ferric chloride in enough water to make 500.0 mL of stock solution. A 50.0 mL aliquot (portion) of this stock solution is then removed and added to 75.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[FeCl3]= 0.0247M
[Fe3+]= 0.0246M
[Cl-]= 0.0738M
3.) A solution is prepared by dissolving 12.80 g of magnesium chlorite in enough water to make 200.0 mL of stock solution. A 25.0 mL aliquot (portion) of this stock solution is then removed and added to 250.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[Mg(ClO2)2]=0.0367M
[Mg2+]= 0.0367M
[ClO2-]= 0.0739M
4.) A solution is prepared by dissolving 25.0 g of aluminum sulfate in enough water to make 125.0 mL of stock solution. A 10.0 mL aliquot (portion) of this stock solution is then removed and added to 15.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[Al2(SO4)3]= 0.234M
[Al3+]= 0.468M
[SO4 2-]= 0.702M
Limiting Reactant
* Ignore the exclamation points (something went wrong)....
1.) A 500.0 mL aliquot of 0.800 M aqueous iron(III) chloride is mixed with ! 400.0 mL of 0.300 M aqueous sodium phosphate.
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
FeCl3 (aq) + Na3PO4 (aq) ----> FePO4(s)+ 3NaCl (aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Fe3+ (aq)+ PO4 3- (aq)----> FePO4(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.4000 mol Fe 3+
0.120 mol PO4 3-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
PO4 3- is the LR
e.)! Calculate the mass of the precipitate produced.
My answer:
18.1g FePO4
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[PO4 3-]= 0M
[Fe3+]= 0.311M
[Cl-]=1.33M
[Na+]= 0.400M
2.) !A 100.0 mL aliquot of 0.200 M aqueous calcium hydroxide is mixed with ! 100.0 mL of 0.200 M aqueous aluminum nitrate.
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
3Ca(OH)2(aq)+2Al(NO3)3(aq)----> 3Ca(NO3)2(aq)+2Al(OH)3(s)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Al3+(aq)+3OH-(aq)----> Al(OH)3(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.0200mol Al3+
0.0400 mol OH-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
OH-is the LR.
e.)! Calculate the mass of the precipitate produced.
My answer:
1.04 g Al(OH)3
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[OH-]= 0M
[Al3+]= 0.0335M
[Ca2+]=0.100M
[NO3-]= 0.300M
3.) !How many grams of solid zinc hydroxide can be prepared by the reaction between ! 100.0 mL of 0.20 M zinc nitrate with 50.0 mL of 0.15 M Ba(OH)2?
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
Zn(NO3)2(aq)+ Ba(OH)2(aq)---> Zn(OH)2(s)+ Ba(NO3)2(aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Zn2+(aq)+2OH-(aq)---> Zn(OH)2(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.020 mol Zn2+
0.015 mol OH-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
OH- is the LR.
e.)! Calculate the mass of the precipitate produced.
My answer:
0.75 g Zn(OH)2
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[OH-]= 0M
[Zn2+]=0.083M
[NO3-]= 0.27M
[Ba2+]= 0.050M
4.) !How many grams of solid barium phosphate can be prepared by the reaction between ! 300.0 mL of 0.50 M barium iodide with 150.0 mL of 0.80 M potassium phosphate?
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
3BaI2(aq)+2K3PO4(aq)----> Ba3(PO4)2(s) + 6KI(aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
3Ba2+(aq)+2PO4 3-(aq)---> Ba3(PO4)2(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.15 mol Ba2+
0.12 mol PO4 3-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
Ba2+ is the LR.
e.)! Calculate the mass of the precipitate produced.
30 g Ba3(PO4)2
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[Ba2+]= 0M
[PO4 3-]= 0.044M
[I-]= 0.67M
[K+]= 0.80M
When a 1.146 gram mixture of solid Na2SO4 and Pb(NO3)2 is added to deionized water, a precipitate of PbSO4 forms and is allowed to settle. From tests preformed on the supernatant, Na2SO4 is determined to be the limiting reactant. After the precipitate is filtered and dried, its mass is determined to be 0.672 grams
a. Write a balanced chemical equation for this reaction. Include correct phases.
My answer:
Na2SO4(s)+Pb(NO3)2(s)----> PbSO4(s)+2NaNO3(aq)
b. Calculate the mass of the limiting reactant that was in the original mixture.
My answer:
0.315g Na2SO4
c. Calculate the mass of the excess reactant that was in the original mixture.
My answer:
0.831 g Pb(NO3)2
d. What is the percent by mass of each salt in the original mixture?
My answer:
27.5% Na2SO4
72.5% Pb(NO3)2
9 answers
2.) A solution is prepared by dissolving 5.00 g of ferric chloride in enough water to make 500.0 mL of stock solution. A 50.0 mL aliquot (portion) of this stock solution is then removed and added to 75.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[FeCl3]= 0.0247M
[Fe3+]= 0.0246M
[Cl-]= 0.0738M
I agree wth [FeCl3]. My answer is 0.02466 which I would round to 0.0247. But that means [Fe^3+] = 0.0247 also.
Then 3*0.02466 = 0.07398 which I would round to 0.0740.
My answer:
[FeCl3]= 0.0247M
[Fe3+]= 0.0246M
[Cl-]= 0.0738M
I agree wth [FeCl3]. My answer is 0.02466 which I would round to 0.0247. But that means [Fe^3+] = 0.0247 also.
Then 3*0.02466 = 0.07398 which I would round to 0.0740.
3.) A solution is prepared by dissolving 12.80 g of magnesium chlorite in enough water to make 200.0 mL of stock solution. A 25.0 mL aliquot (portion) of this stock solution is then removed and added to 250.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[Mg(ClO2)2]=0.0367M
[Mg2+]= 0.0367M
[ClO2-]= 0.0739M
It doesn't make much sense to run all of these since I don't know that we are using the same molar mass. For this one, I have 12.8/159.21 = ?
?/0.200 L = ?
? x (25/275) = 0.036546 which I would round to 0.0366M for [Mg^2+]
and 2*that = 0.07308 which I would round to 0.0731M for [ClO2^-]
One thing I suggest is that you NOT read your calculator at the end of each calculation (and round at that time) but leave that number in the unit and run the next calculation in sequence. If you do that you have, for this case,
(12.80/159.21)x (1/0.500) x (25.0/275.0) = 0.036546 which must be rounded to 3 s.f.
My answer:
[Mg(ClO2)2]=0.0367M
[Mg2+]= 0.0367M
[ClO2-]= 0.0739M
It doesn't make much sense to run all of these since I don't know that we are using the same molar mass. For this one, I have 12.8/159.21 = ?
?/0.200 L = ?
? x (25/275) = 0.036546 which I would round to 0.0366M for [Mg^2+]
and 2*that = 0.07308 which I would round to 0.0731M for [ClO2^-]
One thing I suggest is that you NOT read your calculator at the end of each calculation (and round at that time) but leave that number in the unit and run the next calculation in sequence. If you do that you have, for this case,
(12.80/159.21)x (1/0.500) x (25.0/275.0) = 0.036546 which must be rounded to 3 s.f.
4.) A solution is prepared by dissolving 25.0 g of aluminum sulfate in enough water to make 125.0 mL of stock solution. A 10.0 mL aliquot (portion) of this stock solution is then removed and added to 15.0 mL of water. Calculate the concentrations of all ions in the final solution.
My answer:
[Al2(SO4)3]= 0.234M
[Al3+]= 0.468M
[SO4 2-]= 0.702M
I would have use 0.2338 x 2 = 0.0701 for SO4^2-. The others look ok.
My answer:
[Al2(SO4)3]= 0.234M
[Al3+]= 0.468M
[SO4 2-]= 0.702M
I would have use 0.2338 x 2 = 0.0701 for SO4^2-. The others look ok.
1.) A 500.0 mL aliquot of 0.800 M aqueous iron(III) chloride is mixed with ! 400.0 mL of 0.300 M aqueous sodium phosphate.
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
FeCl3 (aq) + Na3PO4 (aq) ----> FePO4(s)+ 3NaCl (aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Fe3+ (aq)+ PO4 3- (aq)----> FePO4(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.4000 mol Fe 3+I think you're allowed only 3 s.f. here.
0.120 mol PO4 3-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
PO4 3- is the LR
e.)! Calculate the mass of the precipitate produced.
My answer:
18.1g FePO4
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
Except for c, this looks ok down to here,
My answer:
[PO4 3-]= 0M Actually PO4^3- is very small but not zero; however, I don't know if you are to use Ksp to calculate this. If so you must take into consideration that Fe^3+ is a common ion for FePO4 and will decrease the solubility of FePO4.
[Fe3+]= 0.311M
[Cl-]=1.33M
[Na+]= 0.400M
These look ok to me.
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
FeCl3 (aq) + Na3PO4 (aq) ----> FePO4(s)+ 3NaCl (aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Fe3+ (aq)+ PO4 3- (aq)----> FePO4(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.4000 mol Fe 3+I think you're allowed only 3 s.f. here.
0.120 mol PO4 3-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
PO4 3- is the LR
e.)! Calculate the mass of the precipitate produced.
My answer:
18.1g FePO4
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
Except for c, this looks ok down to here,
My answer:
[PO4 3-]= 0M Actually PO4^3- is very small but not zero; however, I don't know if you are to use Ksp to calculate this. If so you must take into consideration that Fe^3+ is a common ion for FePO4 and will decrease the solubility of FePO4.
[Fe3+]= 0.311M
[Cl-]=1.33M
[Na+]= 0.400M
These look ok to me.
.) !A 100.0 mL aliquot of 0.200 M aqueous calcium hydroxide is mixed with ! 100.0 mL of 0.200 M aqueous aluminum nitrate.
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
3Ca(OH)2(aq)+2Al(NO3)3(aq)----> 3Ca(NO3)2(aq)+2Al(OH)3(s)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Al3+(aq)+3OH-(aq)----> Al(OH)3(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.0200mol Al3+
0.0400 mol OH-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
OH-is the LR.
e.)! Calculate the mass of the precipitate produced.
My answer:
1.04 g Al(OH)3
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[OH-]= 0M
[Al3+]= 0.0335M This is the only one I disagree with (slightly). 0.04 mols OH^-/3 = 0.01333 mols AL^3+ used and that subtracted from 0.0200 mols initially leaves 0.006667 Al^3+ and that divided by 0.200 L = 0.00333M
[Ca2+]=0.100M
[NO3-]= 0.300M
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
3Ca(OH)2(aq)+2Al(NO3)3(aq)----> 3Ca(NO3)2(aq)+2Al(OH)3(s)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Al3+(aq)+3OH-(aq)----> Al(OH)3(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.0200mol Al3+
0.0400 mol OH-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
OH-is the LR.
e.)! Calculate the mass of the precipitate produced.
My answer:
1.04 g Al(OH)3
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[OH-]= 0M
[Al3+]= 0.0335M This is the only one I disagree with (slightly). 0.04 mols OH^-/3 = 0.01333 mols AL^3+ used and that subtracted from 0.0200 mols initially leaves 0.006667 Al^3+ and that divided by 0.200 L = 0.00333M
[Ca2+]=0.100M
[NO3-]= 0.300M
3.) !How many grams of solid zinc hydroxide can be prepared by the reaction between ! 100.0 mL of 0.20 M zinc nitrate with 50.0 mL of 0.15 M Ba(OH)2?
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
Zn(NO3)2(aq)+ Ba(OH)2(aq)---> Zn(OH)2(s)+ Ba(NO3)2(aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Zn2+(aq)+2OH-(aq)---> Zn(OH)2(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.020 mol Zn2+
0.015 mol OH-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
OH- is the LR.
e.)! Calculate the mass of the precipitate produced.
My answer:
0.75 g Zn(OH)2
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[OH-]= 0M
[Zn2+]=0.083M
[NO3-]= 0.27M
[Ba2+]= 0.050M
I agree with all of these answers
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
Zn(NO3)2(aq)+ Ba(OH)2(aq)---> Zn(OH)2(s)+ Ba(NO3)2(aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
Zn2+(aq)+2OH-(aq)---> Zn(OH)2(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.020 mol Zn2+
0.015 mol OH-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
OH- is the LR.
e.)! Calculate the mass of the precipitate produced.
My answer:
0.75 g Zn(OH)2
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[OH-]= 0M
[Zn2+]=0.083M
[NO3-]= 0.27M
[Ba2+]= 0.050M
I agree with all of these answers
4.) !How many grams of solid barium phosphate can be prepared by the reaction between ! 300.0 mL of 0.50 M barium iodide with 150.0 mL of 0.80 M potassium phosphate?
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
3BaI2(aq)+2K3PO4(aq)----> Ba3(PO4)2(s) + 6KI(aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
3Ba2+(aq)+2PO4 3-(aq)---> Ba3(PO4)2(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.15 mol Ba2+
0.12 mol PO4 3-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
Ba2+ is the LR.
e.)! Calculate the mass of the precipitate produced.
30 g Ba3(PO4)2
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[Ba2+]= 0M
[PO4 3-]= 0.044M
[I-]= 0.67M
[K+]= 0.80M
I agree with all of these answers.
a.)! Write a complete molecular equation for this reaction, including phases
My answer:
3BaI2(aq)+2K3PO4(aq)----> Ba3(PO4)2(s) + 6KI(aq)
b.)! Write a net ionic equation for this reaction, including phases.
My answer:
3Ba2+(aq)+2PO4 3-(aq)---> Ba3(PO4)2(s)
c.) !For each reacting ion, calculate the number of moles available.
My answer:
0.15 mol Ba2+
0.12 mol PO4 3-
d.) !Use the net ionic equation to determine the limiting reactant.
My answer:
Ba2+ is the LR.
e.)! Calculate the mass of the precipitate produced.
30 g Ba3(PO4)2
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
My answer:
[Ba2+]= 0M
[PO4 3-]= 0.044M
[I-]= 0.67M
[K+]= 0.80M
I agree with all of these answers.
When a 1.146 gram mixture of solid Na2SO4 and Pb(NO3)2 is added to deionized water, a precipitate of PbSO4 forms and is allowed to settle. From tests preformed on the supernatant, Na2SO4 is determined to be the limiting reactant. After the precipitate is filtered and dried, its mass is determined to be 0.672 grams
a. Write a balanced chemical equation for this reaction. Include correct phases.
My answer:
Na2SO4(s)+Pb(NO3)2(s)----> PbSO4(s)+2NaNO3(aq)
I don't know what your instructor has told you to do about these but I wouldn't write this equation this way BECAUSE Na2SO4 solid is not likely to react with Pb(NO3)2 solid. Since you added these to water I would write it as
Na2SO4(s) + H2O ==> Na2SO4(aq)
Pb(NO3)2(s) + H2O ==> Pb(NO3)2(aq), then
Na2SO4(aq) + Pb(NO3)2(aq) ==> etc.
b. Calculate the mass of the limiting reactant that was in the original mixture.
My answer:
0.315g Na2SO4
I calculated 0.317; probably because of a difference of molar masses. I used 303.26 for PbSO4 and 143.043 for Na2SO4 and rounded 0.31697 to 0.317g Na2SO4.
c. Calculate the mass of the excess reactant that was in the original mixture.
My answer:
0.831 g Pb(NO3)2
OK for you 0.315 above.
d. What is the percent by mass of each salt in the original mixture?
My answer:
27.5% Na2SO4
72.5% Pb(NO3)2
OK for your 0.315 answer above.
a. Write a balanced chemical equation for this reaction. Include correct phases.
My answer:
Na2SO4(s)+Pb(NO3)2(s)----> PbSO4(s)+2NaNO3(aq)
I don't know what your instructor has told you to do about these but I wouldn't write this equation this way BECAUSE Na2SO4 solid is not likely to react with Pb(NO3)2 solid. Since you added these to water I would write it as
Na2SO4(s) + H2O ==> Na2SO4(aq)
Pb(NO3)2(s) + H2O ==> Pb(NO3)2(aq), then
Na2SO4(aq) + Pb(NO3)2(aq) ==> etc.
b. Calculate the mass of the limiting reactant that was in the original mixture.
My answer:
0.315g Na2SO4
I calculated 0.317; probably because of a difference of molar masses. I used 303.26 for PbSO4 and 143.043 for Na2SO4 and rounded 0.31697 to 0.317g Na2SO4.
c. Calculate the mass of the excess reactant that was in the original mixture.
My answer:
0.831 g Pb(NO3)2
OK for you 0.315 above.
d. What is the percent by mass of each salt in the original mixture?
My answer:
27.5% Na2SO4
72.5% Pb(NO3)2
OK for your 0.315 answer above.
1 answer