Question

N2(g) + 3H2(g) → 2NH3(g)

5.00 g N2 is reacted with 5.00 g H2

How much reactant is leftover after the reaction?​

Answer 1

Molar mass of N2 = 28.02 g/mol

mass(N2)= 5.0 g

number of mol of N2,

n = mass of N2/molar mass of N2

=(5.0 g)/(28.02 g/mol)

= 0.1784 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 5.0 g

number of mol of H2,

n = mass of H2/molar mass of H2

=(5.0 g)/(2.016 g/mol)

= 2.48 mol

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 0.1784 mol of N2, 0.5353 mol of H2 is required

But we have 2.4802 mol of H2

so, N2 is limiting reagent

we will use N2 in further calculation

According to balanced equation

mol of H2 reacted = (3/1)* moles of N2

= (3/1)*0.1784

= 0.5353 mol

mol of H2 remaining = mol initially present - mol reacted

mol of H2 remaining = 2.4802 - 0.5353

mol of H2 remaining = 1.9448 mol

Molar mass of H2 = 2.016 g/mol

mass of H2,

m = number of mol * molar mass

= 1.945 mol * 2.016 g/mol

= 3.92 g

Answer:

3.92 g of H2 will be leftover