Question

Consider the following BALANCED equation where 80.3 g of NO and 28.6 g of H2 are mixed and reacted:

The Equation: 2NO(g) + 5H₂(g) --> 2NH₃(g) + 2H₂O(g)

a) How many moles of the NH₃ could be produced from the given mass of each reactant?

b) What is the limiting reagent (reactant)?

c) How many grams of NH₃ are theoretically produced?

d) What was the percent yield if only 1.4 g of NH₃ was produced?

e) How much of the excess reagent remains after the reaction?

Answer 1

a)

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

mass(NO)= 80.3 g

number of mol of NO,

n = mass of NO/molar mass of NO

=(80.3 g)/(30.01 g/mol)

= 2.676 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 28.6 g

number of mol of H2,

n = mass of H2/molar mass of H2

=(28.6 g)/(2.016 g/mol)

= 14.19 mol

Balanced chemical equation is:

2 NO + 5 H2 ---> 2 NH3 + 2 H2O

2 mol of NO reacts with 5 mol of H2

for 2.6758 mol of NO, 6.6894 mol of H2 is required

But we have 14.1865 mol of H2

so, NO is limiting reagent

we will use NO in further calculation

According to balanced equation

mol of NH3 formed = (2/2)* moles of NO

= (2/2)*2.6758

= 2.6758 mol

Answer: 2.68 mol

b)

NO is limiting

c)

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

= 2.676*17.03

= 45.58 g

Answer: 45.6 g

d)

% yield = actual mass*100/theoretical mass

= 1.4*100/45.58

= 3.0716%

Answer: 3.07 %

e)

According to balanced equation

mol of H2 reacted = (5/2)* moles of NO

= (5/2)*2.6758

= 6.6894 mol

mol of H2 remaining = mol initially present - mol reacted

mol of H2 remaining = 14.1865 - 6.6894

mol of H2 remaining = 7.4971 mol

Molar mass of H2 = 2.016 g/mol

mass of H2,

m = number of mol * molar mass

= 7.497 mol * 2.016 g/mol

= 15.11 g

Answer: 15.1 g