Question

Balanced the equation for this reaction, and answer the questions.

___ N2 + ___ H2 ➞ ___ NH3

a) If 25.0 g N2 and 18.75 g H2 are used for this reaction, which reactant will be the limiting reactant? Show your

proof.

b) How much of the excess reactant will be left over?

Answer 1

Balanced equation: N2 + 3 H2 ➞ 2NH3.

a) The limiting reactant is H2

b) The excess reactant is N2

Step-by-step explanation:

hydrogen will be the limiting reagent as 3 moles are hydrogen are used up for every one mole of nitrogen to form ammonia

Answer 2

(a) The molar mass of N2 is 28 g/mol, and the molar mass of H2 is 2 g/mol.

n(N2) = 25.0 g / 28 g/mol = 0.893 mol

n(H2) = 18.75 g / 2 g/mol = 9.375 mol

calculate the maximum amount of product that can be formed from each reactant:

From N2: n(NH3) = 0.893 mol N2 x (2 mol NH3 / 1 mol N2) = 1.786 mol NH3

From H2: n(NH3) = 9.375 mol H2 x (2 mol NH3 / 3 mol H2) = 6.25 mol NH3

Therefore, the limiting reactant is N2

(B) For N2: 0.893 mol - 0.893 mol = 0 mol

For H2: 9.375 mol - (0.893 mol x 3 mol H2 / 1 mol N2) = 6.696 mol

Therefore, 6.696 mol of H2 will be left over. To convert this to a mass, we can use the molar mass of H2:

m(H2) = 6.696 mol x 2 g/mol = 13.392 g H2