Chemistry Physical - Week 10 Assessment

Question 1

For limiting reagents, before you decide out of the two which one is limiting, what do you make sure?

Answer 1

You check if there’s a big number in front of any one of the two compunds. If there is, you divide by that number to get real limiting reagent.

Question 2

Calcium sulfide reacts with calcium sulfate as shown.

CaS + 3 CaSO4 → 4 CaO + 4 SO2

2.50 g of calcium sulfide are heated with 9.85 g of calcium sulfate until there is no further reaction.

Show that calcium sulfate is the limiting reagent in this reaction and then calculate mass of sulfur dioxide.

(5)

Answer 2

1) CaS moles = 2.5/72.2 = 0.0346
2) CaSO4 moles = 9.85/136.2 = 0.0723 BUT we need to divide by 3 since it’s 3 CaSO4, which means actual moles = 0.0241, which means CaSO4 is limiting reagent.
3) To get to SO2, multiply 0.0723 (because this is still the number of moles in front of number - we just divide to get limiting reagent) by 4/3 (molar ratio) and get 0.096
4) Multiply this by Mr of SO2 - 0.096 x 64.1 = 6.18 g

Question 3

What is the importance of percentage yield? (1)

Answer 3

Efficient conversion of reactants to products

Question 4

What is the importance of percentage atom economy? (1)

Answer 4

You can maximise rate of reactants

Question 5

5.0 g of an oxide contains 4.0 g of molybdenum. What is the empirical formula of this oxide? (1)

Answer 5

1) 5-4 = 1 g of oxide
2) 1/16 = 0.0625 - Oxygen
3) 4/96 = 0.041 - Molybdenum
4) 0.0625/0.041 = 1: 1.5
5) x2 to get 2:3
6) Which means Mo2O3

Question 6

What is the percentage yield when 20 g of aluminium are produced from 50 g of aluminium oxide? (1)

2 Al2O3 → 4 Al + 3O2

Answer 6

1) 50/102 = **0.49 **(Alumium Oxide moles)
2) 0.49 x 2 = 0.98 (moles of Al)
3) 0.98 x 27 = 26.47 (g of Al)
4) 20/26.47 x 100 ≈ 76%