Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, accord

Question

Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation

3H2(g)+N2(g)2NH3(g)

NOTE: Throughout this tutorial use molar masses expressed to five significant figures.
Part A
How many moles of NH3 can be produced from 18.0 mol of H2 and excess N2?
Part B
How many grams of NH3 can be produced from 4.50 mol of N2 and excess H2.
Part C
How many grams of H2 are needed to produce 12.57 g of NH3?
Part D
How many molecules (not moles) of NH3 are produced from 1.99×104 g of H2?

Explanation / Answer

3H2(g)+N2(g)2NH3(g)

(a)

3 moles of H2 produce 2 moles of NH3

18 mol of H2 produce (2 / 3) * 18 = 12 moles of NH3

(b)

1 mol of N2 produce 2 mol of NH3

4.5 mol of N2 produce 2 * 4.5 = 9 mol of NH3

9 mol of NH3 = 9 mol * 17 g/mol = 153 g of NH3   ( Molar mass of NH3 = 17 g/mol)

(c)

12.57 g of NH3 = 12.57 / 17 mol of NH3

mol of H2 = ( 3 / 2 ) * 12.57 / 17 = 1.109 mol

Mass of H2 = 1.109 mol * 2 g/mol = 2.218 g of H2

(d)

mol of H2 = 1.99 x 10^-4 / 2 = 0.995 x 10^-4 mol

mol of NH3 = (2 / 3) * 0.995 x 10^-4 mol = 0.663 x 10^-4 mol

Molecules of NH3 = 0.663 x 10^-4 mol * 6.022 x 10^23 mol^-1

Molecules of NH3 = 3.994 x 10^19 molecules

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