Question

Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water.

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

If 3.45 g of N2H4 reacts and produces 0.350 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

I got 13.0, and it said that it wasn't correct.

Answer 1

Molar mass of N2H4 = 2*MM(N) + 4*MM(H)

= 2*14.01 + 4*1.008

= 32.052 g/mol

mass of N2H4 = 3.45 g

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

= 3.45/32.052

= 0.1076 mol

we have the Balanced chemical equation as:

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

From balanced chemical reaction, we see that

when 1 mol of N2H4 reacts, 1 mol of N2 is formed

mol of N2 formed = moles of N2H4

= 0.1076 mol

This is theoretical mol of N2 produced

we have:

P = 1.0 atm

V = 0.35 L

T = 295.0 K

find number of moles using:

P * V = n*R*T

1 atm * 0.35 L = n * 0.08206 atm.L/mol.K * 295 K

n = 1.445*10^-2 mol

This is actual mol produced

Now use:

% yield = actual mol * 100 / theoretical mol

= (1.445*10^-2)*100 / (0.1076)

= 13.4 %

Answer: 13.4 %