Question

The Haber Process involves nitrogen gas combining with hydrogen gas to produce ammonia.

N2 + 3H2 → 2NH3

10.0 grams of nitrogen gas is reacted with 10.0 grams of hydrogen gas.

Find the following: the molar mass of reactants and products, the limiting reactant, the excess reactant, the amount of ammonia produced.

Answer 1

N2 = 10g
H2 = 10g

To find the limiting and excess reactants, convert the mass to moles and divide by the coefficient of the reactant, if there is one

moles (n) = mass (m) / molar mass (M)

n(N2) = 10/(2x14) = 0.357 mol
n(H2) = 10/(2x1) = 5 mol = 5/3 = 1.667 mol

Nitrogen gas is the limiting reactant while hydrogen gas is the excess reactant.

You can do the rest following similar steps :)

For the amount of ammonia produced, just do a ratio using the limiting reactant.

Answer 2

Molar masses:

Nitrogen gas (N2): 28.02 g/mol

Hydrogen gas (H2): 2.02 g/mol

Ammonia (NH3): 17.03 g/mol

The number of moles of N2 and H2 are:

N2: 10.0 g / 28.02 g/mol = 0.357 moles

H2: 10.0 g / 2.02 g/mol = 4.95 moles

Since 3 moles of H2 are required for every mole of N2, H2 is the limiting reactant and N2 is the excess reactant.

The amount of ammonia produced can be calculated using the limiting reactant:

0.357 moles N2 * 2 moles NH3/1 mole N2 = 0.714 moles NH3

0.714 moles NH3 * 17.03 g/mol = 12.28 g NH3.

Therefore, 10.0 grams of nitrogen gas reacts with 10.0 grams of hydrogen gas to produce 12.28 grams of ammonia.