Ammonia can be synthesized by the reaction: 3H2(g) + N2(g) —-> 2NH3(g) What is t

Question

Ammonia can be synthesized by the reaction:
3H2(g) + N2(g) —-> 2NH3(g)
What is the theoretical yield of ammonia, in kg, that we can synthesize from 5.43kg of H2 and 32.4 kg of N2? Ammonia can be synthesized by the reaction:
3H2(g) + N2(g) —-> 2NH3(g)
What is the theoretical yield of ammonia, in kg, that we can synthesize from 5.43kg of H2 and 32.4 kg of N2?
3H2(g) + N2(g) —-> 2NH3(g)
What is the theoretical yield of ammonia, in kg, that we can synthesize from 5.43kg of H2 and 32.4 kg of N2?

Explanation / Answer

Molar mass of H2 = 2.016 g/mol

mass of H2 = 5.43 Kg = 5430 g

we have below equation to be used:

number of mol of H2,

n = mass of H2/molar mass of H2

=(5430.0 g)/(2.016 g/mol)

= 2.693*10^3 mol

Molar mass of N2 = 28.02 g/mol

mass of N2 = 32.4 Kg = 32400 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(32400.0 g)/(28.02 g/mol)

= 1.156*10^3 mol

we have the Balanced chemical equation as:

3 H2 + N2 ---> 2 NH3

3 mol of H2 reacts with 1 mol of N2

for 2693.4524 mol of H2, 897.8175 mol of N2 is required

But we have 1156.3169 mol of N2

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)*2693.4524

= 1.796*10^3 mol

we have below equation to be used:

mass of NH3 = number of mol * molar mass

= 1.796*10^3*17.03

= 3.059*10^4 g

= 30.6 Kg

Answer: 30.6 Kg

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