Question

Consider the unbalance equation :

N2H4 + H2O2 = N2 + H20

What is the yield of the above reaction if 1.0 g of N2H4 was reacted if 1.0 g of H20 and 1 g of N2 was isolated

Answer 1

N₂H₄ + H₂O₂ N₂ + H₂O

The balance equation is

N₂H₄ + 2 H₂O₂ N₂ + 4 H₂O

In the above reaction equation;

1 mole of N₂H₄ produces 1 mole of N₂ and 4 moles of H₂O

Now:

1.0 g of N2H4
Molar mass of N2H4 = 32 g/mol
So, 32 g of N2H4 = 1 mol
1 g of N2H4 = (1/32) mol = 0.0313 mol

1.0 g of H2O
Molar mass of H2O = 18 g/mol
So, 18 g of H2O = 1 mol
​ 1 g of H2O = (1/18) mol = 0.056 mol

1.0 g of N2
Molar mass of N2 = 28 g/mol
So, 28 g of N2 = 1 mol
​ 1 g of N2 = (1/28) mol = 0.0357 mol

Again,

1 mole of N₂H₄ produces 1 mole of N₂ and 4 moles of H₂O

0.0313 mol of N₂H₄ produces 0.0313 mol of N₂ and 4 (0.0313 mol) of H₂O

0.0313 mol of N₂H₄ produces 0.0313 mol of N₂ and 0.1252 mol of H₂O.

Since 1 g of N2 = 0.0357 mol > expetected mol (0.0313 mol). So, N2 mass is discarded.

Now, considering H2O.

Molar mass of H2O = 18 g/mol
So, 1 mol of H2O = 18 g
​ 0.1252 mol of H2O = 0.1252 x 18 g = 2.25 g (theoritical yield)

Practical Yield= 1 g

% Yield = (1 g / 2.25 g) x 100
= 44.44 %