Question

hydrazine N2H4 reacts with oxygen to form nitrogen gas and water

Hydrazine, N, H, reacts with oxygen to form nitrogen gas and water. N,H, (aq) +0,6) N,(g) + 2 H2O(1) If 2.45 g of N, H, reacts with excess oxygen and produces 0.950 L of N,, at 295 K and 1.00 atm, what is the percent yield of the reaction? percent yield:

Answer 1

Step 1: calculate theoretical yield in mol

Molar mass of N2H4,

MM = 2*MM(N) + 4*MM(H)

= 2*14.01 + 4*1.008

= 32.052 g/mol

mass of N2H4 = 2.45 g

mol of N2H4 = (mass)/(molar mass)

= 2.45/32.05

= 7.644*10^-2 mol

According to balanced equation

mol of N2 formed = moles of N2H4

= 7.644*10^-2 mol

Step 2: calculate actual yield in mol

Given:

P = 1.0 atm

V = 0.95 L

T = 295.0 K

find number of moles using:

P * V = n*R*T

1 atm * 0.95 L = n * 0.08206 atm.L/mol.K * 295 K

n = 3.924*10^-2 mol

Step 3:

% yield = actual yield * 100 / theoretical yield

= (3.924*10^-2)*100 / (7.644*10^-2)

= 51.3 %

Answer: 51.3 %