Question

Nitrogen gas and hydrogen gas react to form ammonia gas, NH3. If you reacted 10.0kg of nitrogen gas with an excess of hydrogen, how many grams of ammonia could be produced? If 10.5kg were actually produced in your experiment, calculate your percent yield.

Answer 1

1)

Molar mass of N2 = 28.02 g/mol

mass(N2)= 10 Kg = 10000 g

number of mol of N2,

n = mass of N2/molar mass of N2

=(10000 g)/(28.02 g/mol)

= 3.569*10^2 mol

we have the Balanced chemical equation as:

N2 + 3 H2 ---> 2 NH3

From balanced chemical reaction, we see that

when 1 mol of N2 reacts, 2 mol of NH3 is formed

mol of NH3 formed = (2/1)* moles of N2

= (2/1)*356.8879

= 713.7759 mol

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = number of mol * molar mass

= 7.138*10^2*17.03

= 1.216*10^4 g

=12.2 Kg

Answer: 12.2 Kg

2)

% yield = actual mass*100/theoretical mass

= 10.5*100/12.2

= 86.4 %

Answer: 86.4 %