Question

For the chemical reaction shown, 2H2O2(l)+N2H4(l)⟶4H2O(g)+N2(g) determine how many grams of N2 are produced from the reaction of 9.55g of H2O2 and 5.81 g of N2H4.

Answer 1

Number of moles of H2O2 = 9.55 g / 34.0147 g/mol = 0.281 mol

Number of moles of N2H4 = 5.81 g / 32.0452 g/mol = 0.181 mol

From the balanced equation we can say that

2 mole of H2O2 requires 1 mole of N2H4 so

0.281 mole of H2O2 will require

= 0.281 mole of H2O2 *(1 mole of N2H4 / 2 mole of H2O2)

= 0.141 mole of N2H4

But we have 0.181 mole of N2H4 which is in excess so N2H4 is an excess reactant and H2O2 is limiting reactant

From the balanced equation we can say that

2 mole of H2O2 produces 1 mole of N2 so

0.281 mole of H2O2 will produce

= 0.281 mole of H2O2 *(1 mole of N2 / 2 mole of H2O2)

= 0.141 mole of N2

mass of 1 mole of N2 = 14.0067 g so

the mass of 0.141 mole of N2 = 1.97 g

Therefore, the mass of N2 produced would be 1.97 g