Asked by Anonymous
a)How many calcium ions are in 2.17 moles of calcium phosphate?
b)If 1.41 g of zinc or attitude 1.85 g of hydrochloric acid identify the limiting and excess reagents give reasons for your choice and determine how many moles of the excess reagent remains
c)What volume of 0.125 mol/L sodium hydroxide solution is required to completely react with 15.0 mL of 0.100 mol/L silver sulfate? give your answer in millilitres
b)If 1.41 g of zinc or attitude 1.85 g of hydrochloric acid identify the limiting and excess reagents give reasons for your choice and determine how many moles of the excess reagent remains
c)What volume of 0.125 mol/L sodium hydroxide solution is required to completely react with 15.0 mL of 0.100 mol/L silver sulfate? give your answer in millilitres
Answers
Answered by
DrBob222
a)How many calcium ions are in 2.17 moles of calcium phosphate?
<b>2.17 mols Ca3(PO4)2 has 2.17 x 3 = 6.51 mols Ca ions. 1 mole of any ion contains 6.02E23 individual ions.</b>
b)If 1.41 g of zinc (you must have meant "are added to) 1.85 g of hydrochloric acid identify the limiting and excess reagents give reasons for your choice and determine how many moles of the excess reagent remains
<b> Zn + 2HCl ==> ZnCl2 + H2
mols Zn = g/atomic mass = 1.41/65.4 = 0.021 estimated. You need to redo it.
mols HCl = 1.85/36.5 = estimated 0.05.
How many mols H2 could each produce? For Zn that's
0.021 mols H2. For HCl that's 0.05 x 1/2 = 0.025. Therefore, the smaller amount will be formed; i.e., 0.021 mols H2 will form, Zn is the limiting reagent (LR) and the Excess reagent (ER) is HCl.
If we used all 0.021 mols Zn, we will use 2*0.021 = 0.042 mols HCl and 0.05 - 0.042 = ? mols HCl, the ER, remain un-reacted. </b>
c)What volume of 0.125 mol/L sodium hydroxide solution is required to completely react with 15.0 mL of 0.100 mol/L silver sulfate? give your answer in millilitres.
<b> You don't show a reaction, The following is just ONE that can be written.
2NaOH + Ag2SO4 ==> 2AgOH +Na2SO4. The AgOH is not stable but I'll leave it at that.
mols Ag2SO4 = M x L = 0.1 x 0.015 L = 0.0015
0.0015 mols Ag2SO4 will required twice that or 0.0030 mols NaOH. Then M mols/L. You know mols and M of NaOH, substitute and solve for L and convert to mL.
Post your work if you run into a problem.
<b>2.17 mols Ca3(PO4)2 has 2.17 x 3 = 6.51 mols Ca ions. 1 mole of any ion contains 6.02E23 individual ions.</b>
b)If 1.41 g of zinc (you must have meant "are added to) 1.85 g of hydrochloric acid identify the limiting and excess reagents give reasons for your choice and determine how many moles of the excess reagent remains
<b> Zn + 2HCl ==> ZnCl2 + H2
mols Zn = g/atomic mass = 1.41/65.4 = 0.021 estimated. You need to redo it.
mols HCl = 1.85/36.5 = estimated 0.05.
How many mols H2 could each produce? For Zn that's
0.021 mols H2. For HCl that's 0.05 x 1/2 = 0.025. Therefore, the smaller amount will be formed; i.e., 0.021 mols H2 will form, Zn is the limiting reagent (LR) and the Excess reagent (ER) is HCl.
If we used all 0.021 mols Zn, we will use 2*0.021 = 0.042 mols HCl and 0.05 - 0.042 = ? mols HCl, the ER, remain un-reacted. </b>
c)What volume of 0.125 mol/L sodium hydroxide solution is required to completely react with 15.0 mL of 0.100 mol/L silver sulfate? give your answer in millilitres.
<b> You don't show a reaction, The following is just ONE that can be written.
2NaOH + Ag2SO4 ==> 2AgOH +Na2SO4. The AgOH is not stable but I'll leave it at that.
mols Ag2SO4 = M x L = 0.1 x 0.015 L = 0.0015
0.0015 mols Ag2SO4 will required twice that or 0.0030 mols NaOH. Then M mols/L. You know mols and M of NaOH, substitute and solve for L and convert to mL.
Post your work if you run into a problem.
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