Question

Assume that you have 1.48 mol of H2 and 3.47 mol of N2. How many grams of ammonia (NH3)can you make, and how many grams of which reactant will be left over? 3H2+N2→2NH3

Part A mNH3 =   g   SubmitMy AnswersGive Up Part B mH2 =   g   SubmitMy AnswersGive Up Part C mN2 =   g   SubmitMy AnswersGive Up

Answer 1

A)

we have the Balanced chemical equation as:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 3.47 mol of N2, 10.41 mol of H2 is required

But we have 1.48 mol of H2

so, H2 is limiting reagent

we will use H2 in further calculation

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

From balanced chemical reaction, we see that

when 3 mol of H2 reacts, 2 mol of NH3 is formed

mol of NH3 formed = (2/3)* moles of H2

= (2/3)*1.48

= 0.9867 mol

we have below equation to be used:

mass of NH3 = number of mol * molar mass

= 0.9867*17.03

= 16.8 g

Answer: 16.8 g

B)

Answer: 0 g

because H2 is limiting reagent

C)

From balanced chemical reaction, we see that

when 3 mol of H2 reacts, 1 mol of N2 is formed

mol of N2 reacted = (1/3)* moles of H2

= (1/3)*1.48

= 0.4933 mol

mol of N2 remaining = mol initially present - mol reacted

mol of N2 remaining = 3.47 - 0.4933

mol of N2 remaining = 2.9767 mol

Molar mass of N2 = 28.02 g/mol

we have below equation to be used:

mass of N2,

m = number of mol * molar mass

= 2.977 mol * 28.02 g/mol

= 83.4 g

Answer: 83.4 g