stoichiometry question

Question

A 1.81g sample of ZnS was reacted with 25.0 mL of a nitric acid solution that had a concentration of 0.836 mol/L. One of the products of this reaction is Sulphur. It was collected at the end of the experiment and was found to have a mass of 0.173 g. A second product was water vapour. This was also collected. The lab had a temperature of 305 K and a pressure of 99.4 kPa on the day the experiment was performed.
The balanced equation for the reaction that occured is:
4 ZnS(s) + 10 HNO3 (aq) --> 4 S(s) + 3 H2O(g) + 4 Zn(NO3)2 (aq) + NH4NO3 (aq)

a) Determine the volume of H₂O vapour in millilitres that was obtained in the reaction

b) Calculate the percentage yield of S(s)

Answer 1

Hi Allan,

 

4ZnS(s) + 10HNO₃(aq) ----> 4S(s) +  3H₂O(g)  + 4Zn(NO₃)₂(aq) + NH₄NO₃(aq)

4mole         10 moles           4 moles   3 moles      4 moles             1 mole

 

1.81g         25mL                 0.173g

                 0.836mol/L

(a) 

Number of moles of HNO₃ = (0.836 mol/L)(25mL *1L/1000mL) =0.0209 mole

In the equation, 10 moles of nitric acid resulted in 3 moles of water

Number of moles of water = (3 mole H₂O/10 mole HNO₃)(0.0209 mole HNO₃)

                                      =0.00627 mole

From Ideal gas law

PV = nRT ; T = 305 K, P = 99.4kPa, V =?, n = 0.00627 mole, R = 8.314 kPa L mol^–1 K^–1.

V = nRT/P

   =(0.00627 mol)(8.314kPa L mol^-1 K^-1)(305 K)/99.4 kPa

   =0.15665 L

The volume of water vapour obtained in the reaction is 159.95 mL

(b) ZnS(s) has a molecular mass = 65.38 +32.06 =97.44g/mole

Number of moles reacted = 1.81g/97.44g/mole = 0.01858 mole

Since 4 moles of ZnS(s) produces 4 moles of S(s), then

number of moles of S(s) produced= 0.01858 mole

atomic mass of S(s) = 32.06 g/mole

Ideally, the amount of S(s) produced = (32.06 g/mole)(0.01858 mole) =0.596g

Percentage yield = (actual amount/theoretical amount)x100% = (0.173g/0.596)x100% =29.0%

The percentage yield of S(s) is 29.0%