Solve chemistry study guide.

Question

3. Suppose that you have 110.0 mL of Na2CO3 that is 2.5 M. If this breaks down according to the following chemical equation, how many liters of CO2 gas will be released at STP?

I assume that for this question I am using R=0.08206(L*atm)/(K*Mol)

7. In the following reaction, 250.0 g of solid zinc reacts with 1.50 L of aqueous copper (II) sulfate that is 2.80 M. How many grams of zinc sulfate are produced?

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

Answer 1

3.  Any time you see STP, you don't have to use PV = nRT nor do you have to use R.  Remember that at STP, 1 mole of any gas = 22.4 liters.  So, as soon as you see STP in a problem, automatically think of this value (22.4 L/mole).

You didn't provide the decomposition reaction and there are several ways in which Na₂CO3 can decompose. Since you specify a solution, let us assume the decomposition reaction is Na₂CO₃ + H₂O ==> H₂O + CO₂ (+NaOH)

moles Na₂CO₃ = 0.1100 L x 2.5 mol/L = 0.275 moles Na₂CO₃

moles CO₂ formed = 0.275 moles Na₂CO₃ + 1 mole CO₂/mole Na₂CO₃ = 0.275 moles CO₂

Volume of CO₂ at STP = 0.275 moles x 22.4 L/mole = 6.16 liters = 6.2 liters (to 2 significant figures).

 

7.  Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)

moles Zn(s) present = 250.0 g Zn x 1 mole Zn/65.38 g = 3.824 moles Zn

moles CuSO₄ present = 1.50 L x 2.80 mol/L = 4.20 moles

Which reactant is limiting?  This would be Zn because of the 1 : 1 mole ratio of Zn to CuSO₄.

Moles ZnSO4 that can be produced will be limited by the moles of Zn present

Thus...

3.824 moles Zn x 1 mole ZnSO₄/mole Zn = 3.824 moles of ZnSO₄ can be produced

mass of ZnSO₄ = 3.824 moles x 161.5 g/mole = 618 moles (to 3 significant figures)