How do you calculate the volumes of the acid solutions (1.50M HCl, 1.50M HNO3, and 1.50 M H2SO4?
3 answers
I don't understand the question. 1.50M HCl = 1.5 moles/L of solution. The volumes are depended upon the reaction and you have given no reaction.
All of those are separate reactions that are being reacted with 50 mL of 1.50 M NH3.
The concept is the same for all.
1. Write and balance the equation.
NH3 + HCl ==> NH4Cl
2. Calculate moles NH3.
moles NH3 = M x L
3. Using the coefficients in the balanced equation, convert moles NH3 to moles of the acid.
0.075 moles NH3 x (1 mole HCl/1 mole NH3) = 0.075 x (1/1) = 0.075 moles HCl
4. Convert moles HCl to volume.
M = moles/L. YOu have moles and M, solve for L.
Same concept for H2SO4 but it is a little different in step 3.
2NH3 + H2SO4 ==> (NH4)2SO4
moles NH3 = M x L = 0.075
moles H2SO4 = 0.075 moles NH3 x (1 mole H2SO4/2 moles NH3) = 0.075 x (1/2) = ??
Then M = moles/L.
The whole point of the problem is to illustrate two points.
a. When the SAME concns and are used, the VOLUME is the same for 1:1 reactions such as HCl and HNO3 but
b. they are not the same, even for the same concn, if the ratio is not 1:1 (as in the H2SO4 part).
1. Write and balance the equation.
NH3 + HCl ==> NH4Cl
2. Calculate moles NH3.
moles NH3 = M x L
3. Using the coefficients in the balanced equation, convert moles NH3 to moles of the acid.
0.075 moles NH3 x (1 mole HCl/1 mole NH3) = 0.075 x (1/1) = 0.075 moles HCl
4. Convert moles HCl to volume.
M = moles/L. YOu have moles and M, solve for L.
Same concept for H2SO4 but it is a little different in step 3.
2NH3 + H2SO4 ==> (NH4)2SO4
moles NH3 = M x L = 0.075
moles H2SO4 = 0.075 moles NH3 x (1 mole H2SO4/2 moles NH3) = 0.075 x (1/2) = ??
Then M = moles/L.
The whole point of the problem is to illustrate two points.
a. When the SAME concns and are used, the VOLUME is the same for 1:1 reactions such as HCl and HNO3 but
b. they are not the same, even for the same concn, if the ratio is not 1:1 (as in the H2SO4 part).