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Chapter 3: Problem 123

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
The maximum mass of ammonia that can be produced is \(1215.37 \ \mathrm{g}\), and \(283.71 \ \mathrm{g}\) of hydrogen will remain unreacted.

Step by step solution

01

Identify the chemical equation and conversion factors

We are given the balanced chemical equation for the synthesis of ammonia: \(\mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightarrow 2\mathrm{NH}_3(g)\) We are also given the initial masses of the reactants: \(1.00\times10^{3} \ \mathrm{g} \ \mathrm{N}_2\) and \(5.00\times10^{2} \ \mathrm{g} \ \mathrm{H}_2\). Using the molar masses of the elements (N: 14.01 g/mol, H: 1.01 g/mol), we can find the moles of each reactant.
02

Calculate the moles of N₂ and H₂

Convert the mass of each reactant to moles: Moles of N₂ = \( \frac{1.00\times10^{3} \ \mathrm{g} \ \mathrm{N}_2}{28.02 \ \mathrm{g \ mol^{-1}}} = 35.69 \ \mathrm{moles \ N}_2\) Moles of H₂ = \( \frac{5.00\times10^{2} \ \mathrm{g} \ \mathrm{H}_2}{2.02 \ \mathrm{g \ mol^{-1}}} = 247.52 \ \mathrm{moles \ H}_2\)
03

Determine the limiting reactant

To find which reactant will be consumed first (limiting reactant), we will calculate the mole ratio for both reactants: Mole ratio of N₂:H₂ = \(\frac{35.69 \ \mathrm{moles \ N}_2}{247.52 \ \mathrm{moles \ H}_2} = 0.144\) Since the mole ratio is less than the required \(1:3\) stoichiometric ratio, nitrogen is the limiting reactant. The reaction will stop when all of the nitrogen is consumed.
04

Calculate the maximum mass of ammonia produced

Based on the limiting reactant (N₂), we can determine the maximum amount of ammonia produced using the stoichiometry of the balanced equation: Moles of NH₃ = 2 × moles of N₂ = 2 × 35.69 = 71.38 moles Now, we can convert the moles of ammonia back into mass: Mass of NH₃ = 71.38 moles × 17.03 g/mol = 1215.37 g So, the maximum mass of ammonia that can be produced is 1215.37 g.
05

Calculate the mass of unreacted starting material

Only hydrogen will remain unreacted. To calculate the mass of hydrogen left, we first need to find out how much hydrogen was consumed: Moles of H₂ consumed = 3 × moles of N₂ = 3 × 35.69 = 107.07 moles Now, we can find the remaining moles of hydrogen and convert it back into mass: Moles of H₂ remaining = 247.52 moles - 107.07 moles = 140.45 moles Mass of H₂ remaining = 140.45 moles × 2.02 g/mol = 283.71 g Thus, 283.71 g of hydrogen will remain unreacted. In conclusion, the maximum mass of ammonia that can be produced is 1215.37 g, and 283.71 g of hydrogen will remain unreacted.

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Most popular questions from this chapter

A 2.077 -g sample of an element, which has an atomic mass between 40 and \(55,\) reacts with oxygen to form 3.708 g of an oxide. Determine the formula of the oxide (and identify the element).

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Hexamethylenediamine $\left(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{N}_{2}\right)$ is one of the starting materials for the production of nylon. It can be prepared from adipic acid $\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\right)$ by the following overall equation: $$ \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}(l)+2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{6} \mathrm{H}_{16} \mathrm{N}_{2}(l)+4 \mathrm{H}_{2} \mathrm{O}(l) $$ What is the percent yield for the reaction if 765 g of hexamethylenediamine is made from \(1.00 \times 10^{3} \mathrm{g}\) of adipic acid?
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