Question

Hydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water. N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 2.45 g of N2H4 reacts with excess oxygen and produces 0.450 L of N2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

Answer:

24.15%

Explanation:

According to the given situation the computation of the percent yield of the reaction is shown below:-

PV = NRT = N = $\frac{PV}{RT}$

Mole of $N_2$ = $\frac{PV}{RT}$

= $\frac{1\times 0.450}{0.0821\times 295}$

= $\frac{0.450}{24.2195}$

= 0.0186

Mole of $N_2H_4 = \frac{2.45}{32}$

= 0.077

Now, the percentage of yield is

= $\frac{Practical\ yield}{Theoretical\ yield}\times 100$

= $\frac{0.0186}{0.077}\times 100$

= 24.15%

Therefore for computing the percentage of yield we simply divide the practical yield by theoretical yield and multiply with 100 so that we can get the result into the percentage form.