grade 12 university chemistry 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 HNO_{3 (aq)} --> 4 S_(s) + 3 H₂O_(g) + 4 Zn(NO₃)_{2 (aq)} + NH₄NO_{3 (aq)}

 

 Calculate the percentage yield of S_(s) 

Answer 1

Find the Limiting Reagent

 

#moles of S produced from ZnS=(#moles ZnS)*(4 moles S/4 moles ZnS)

 

#moles of S produced from HNO₃=(#moles HNO₃)*(4 moles S/10 moles HNO₃)

 

#moles ZnS available=1.81 g*(1mole/97.456g) or 0.0196 mole ZnS

 

#moles HNO₃ available=(0.836 moles/L)*0.025 L or 0.0209 mole HNO₃

 

Find the possible # moles of S from each reagent

 

#moles of S produced from ZnS=(0.0196 mole ZnS)*(4 moles S/4 moles ZnS) or 0.0196 mole S.

 

#moles of S produced from HNO₃=(0.0209 mole HNO₃)*(4 moles S/10 moles HNO₃) or 0.00836 mole S

 

HNO₃ is the Limiting Reagent.  0.00836 mole S will be produced.

 

#grams of S produced=0.00836 mole S*(32.066 g/1 mole S) or

0.268 g S

 

If actual mass of S collected was 0.173 g,

then %yield=(0.173/0.268)*100%

 

64.5%