Given the following equation: 8Fe+S8=8FeS

a. If 4.2 moles of sulfur react, what is the mass of FeS produced?

b. if 13 moles of Fe react, what is the mass of FeS produced?

a. 2954g of FeS

b. 1143g of FeS

Explanation:

Based on the reaction:

8 Fe + S₈ → 8 FeS

8 moles of Fe reacts with 1 mole of S₈ to produce 8 moles of FeS

a. 4.2mol sulfur produce:

4.2mol S₈ × ( 8 mol FeS / 1 mol S₈) = 33.6mol FeS. As molar mass of FeS is 87.92g/mol, mass is:

33.6mol FeS × ( 87.92g / 1 mol FeS) = 2954g of FeS

b. 13mol Fe produce:

13mol Fe × ( 8 mol FeS / 8 mol Fe) = 13mol FeS. In mass:

13 mol FeS × ( 87.92g / 1 mol FeS) = 1143g of FeS

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.