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Chapter 3: Problem 109

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{~g} \mathrm{~N}_{2}\) and \(5.00 \times 10^{2} \mathrm{~g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Short Answer

Expert verified
a) The maximum mass of ammonia produced is \(1215.36\,\mathrm{g}\). b) The mass of unreacted hydrogen is \(283.71\,\mathrm{g}\).

Step by step solution

01

Convert grams of reactants to moles

First, we need to convert grams of both nitrogen (N\(_2\)) and hydrogen (H\(_2\)) to moles using their respective molar masses. The molar mass of N\(_2\) is 28.02 g/mol, and the molar mass of H\(_2\) is 2.02 g/mol. Moles of N\(_2\) = \(\frac{1.00 \times 10^3 g}{28.02\,\mathrm{g/mol}} = 35.69\,\mathrm{mol}\) Moles of H\(_2\) = \(\frac{5.00 \times 10^2 g}{2.02\,\mathrm{g/mol}} = 247.52\,\mathrm{mol}\)
02

Determine the limiting reactant

We can now find the stoichiometric ratio of hydrogen to nitrogen according to the balanced equation: 1 mol N\(_2\): 3 mol H\(_2\) Calculate how much hydrogen is needed for complete consumption of 35.69 mol of nitrogen: Required mol of H\(_2\) = 35.69 mol \(\times\) \(\frac{3\, \mathrm{mol \, H_2}}{1\, \mathrm{mol\, N_2}} = 107.07\, \mathrm{mol}\) Since we have more moles of H\(_2\) than required (247.52 mol > 107.07 mol), nitrogen is the limiting reactant, and hydrogen is in excess.
03

Calculate the maximum mass of NH\(_3\) produced

Now, we can determine the maximum amount of ammonia that can be produced from the limiting reactant, nitrogen: Moles of NH\(_3\) produced = 35.69 mol\(\, \mathrm{N_2}\) \(\times\) \(\frac{2\, \mathrm{mol\, NH_3}}{1\, \mathrm{mol\, N_2}} = 71.38\, \mathrm{mol}\) The molar mass of NH\(_3\) is 17.03 g/mol. Mass of NH\(_3\) = 71.38 mol NH\(_3\) \(\times\) \(17.03\,\mathrm{g/mol} = 1215.36\,\mathrm{g}\) So, the maximum mass of ammonia that can be produced is 1215.36 g.
04

Calculate the mass of unreacted starting material

To determine the mass of unreacted hydrogen, we need to calculate how many moles of hydrogen are consumed in the reaction and convert it back to grams. Moles of H\(_2\) consumed = 35.69 mol N\(_2\) \(\times\) \(\frac{3\, \mathrm{mol\, H_2}}{1\, \mathrm{mol\, N_2}} = 107.07\, \mathrm{mol}\) Moles of H\(_2\) unreacted = Total moles of H\(_2\) - Moles of H\(_2\) consumed = 247.52 mol - 107.07 mol = 140.45 mol Mass of H\(_2\) unreacted = 140.45 mol \(\times\) \(2.02\, \mathrm{g/mol} = 283.71\,\mathrm{g}\) Therefore, 283.71 g of hydrogen remains unreacted.
05

Answers:

a) The maximum mass of ammonia produced is 1215.36 g. b) The mass of unreacted hydrogen is 283.71 g.

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