## Answers

The balanced reaction is as follows:

N2H4(aq) + O2(g) N2(g) + 2H2O(l)

Mass of N2H4 = **3.05 g**

Temperature(T) = **295 K**

Pressure(P) = **1.00 atm**

**Determine the moles of N2H4 as follows:**

Molar mass of N2H4 = **32.0452 g/mol**

= 3.05 g N2H4 x ( 1 mol N2H4 / 32.0452 g N2H4)

= 0.09518 mol N2H4

Use the moles of N2H4 and the mole ratio from the balanced chemical reaction. Determine the moles of N2 as follows:

= 0.09518 mol N2H4 x ( 1 mol N2 / 1 mol N2H4)

**= 0.09518 mol N2**

The ideal gas law equation is as follows:

**PV = nRT**

Rearrange the formula for volume as follows:

V = nRT /P

V = (0.09518 mol x 0.08206 L.atm/mol.K x 295 K) / 1 atm

V = 2.304 L

**Thus, the theoretical yield of N2 gas is 2.304 L**

**Determine the percentage yield as follows:**

Experimental yield of N2 = 0.950 L

Theoretical yield of N2 = 2.304 L

Percent yield = [ Experimetnal yield / Theoretical yield] x 100 %

Percent yield = [ 0.950 L / 2.304 L] x 100 %

**Percent yield = 41.2 %**

**Thus, the percent yield of the reaction will be 41.2 % [3 S.F].**