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Ammonia can also be synthesized by the reaction 3h2(g)+n2(g)→2nh3(g) what is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.22 kg of h2 and 31.5 kg of n2? express the mass in kilograms to three significant figures.

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barnuts

We are given the complete reaction:

3 H2(g)  +  N2(g) → 2 NH3(g)

 

First let us convert mass to moles.

moles H2 = 5.22 kg / (2 kg/kmol) = 2.61 kmol H2

moles N2 = 31.5 kg / (28 kg/kmol) = 1.125 kmol N2

 

Then we find for the limiting reactant. The limiting reactant is the one who has lower (moles/coefficient) ratio.

H2 = 2.61 / 3 = 0.87

N2 = 1.125 / 1 = 1.125

 

Hence the H2 is the limiting reactant so we should base the calculation of NH3 from it. We see that 2 moles of NH3 is produced for every 3 moles of H2, therefore:

moles NH3 = 2.61 kmol H2 * (2 kmol NH3 / 3 kmol H2) = 1.74 kmol

 

The molar mass of NH3 is 17 kg/kmol, therefore the mass NH3 is:

mass NH3 = 1.74 kmol * 17 kg/kmol

mass NH3 = 29.58 kg

meerkat18
Given:
3H2(g)+H2(g)→2NH3(g)

5.22 kg of H2
31.5 kg of N2

Required: Theoretical Yield of NH3

Solution:

Check for the limiting reactant,
5.22 kg H2 x mol/2 kg H2 = 2.61 mol H2
31.5 kg N2 x mol/28 kg N2 = 1.125 mol N2

According to the balanced equation, we need 3 moles of H2 and 1 mole of N2 to produce 2 moles of NH3.  Based on our previous calculation, the limiting reactant is H2 since we have only 2.61 moles.

Solving,

2.61 mols H2 x 2 mols NH3/3 mols H2 = 1.74 mols NH3
1.74 mols NH3 x 17 kg NH3/1 mol NH3 = 29.58 kg NH3

ANSWER: Theoretical Yield is 29.58 kg NH3

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