Part A: How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2? Part B: How many grams of NH3 can be produced from 3.60 mol of N2 and excess H2?

Part C: How many grams of H2 are needed to produce 11.76g of NH3?

Part D: How many molecules (not moles) of NH3 are produced from 8.86 * 10^-4g of H2?

a) 7.0 moles of NH3

b) 61.2 g of NH3

c) 4.15 g of H2

d) 8.9 ×10^19 molecules

Explanation:

Equation of the reaction;

N2(g) + 3H2(g) ⇄NH3(g)

a)

If 3 moles of H2 yields 1 mole of NH3

21 moles of H2 will yield 21 × 1 /3 = 7.0 moles of NH3

b)

1 mole of N2 yields 17 g of NH3

3.6 moles of N2 yields 3.6 moles × 17 g of NH3 = 61.2 g of NH3

c)

If 6g of H2 produces 17 g of NH3

xg of H2 will produce 11.76 g of NH3

x= 6 × 11.76/17

x= 4.15 g of H2

d)

If 6g of hydrogen yields 6.02 × 10^23 molecules of NH3

8.86 × 10^-4g of H2 yields 8.86 × 10^-4g × 6.02 × 10^23 /6 = 8.9 ×10^19 molecules

dynamic equilibrium is the answer.