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Chapter 3: Problem 117

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 \(1.22\times 10^3~g~NH_{3}\). The mass of the leftover starting material, hydrogen gas, is approximately 283.6 g.

Step by step solution

01

Calculate the moles of the starting materials

First, we need to convert the mass of nitrogen and hydrogen gases into moles. To do this, we will use the molar mass of each gas. The molar mass of N₂ is 28.02 g/mol and H₂ is 2.02 g/mol. Nitrogen gas: \( \frac{1.00 \times 10^3 g~N_{2}}{28.02 g~N_{2}/mol~N_{2}} = 35.7 mol~N_{2} \) Hydrogen gas: \( \frac{5.00 \times 10^2 g~H_{2}}{2.02 g~H_{2}/mol~H_{2}} = 247.5 mol~H_{2} \)
02

Determine the limiting reactant

Using the stoichiometry of the balanced equation, we can determine which reactant is the limiting reactant. In this reaction, one mole of N₂ reacts with three moles of H₂ to produce two moles of NH₃. To find the limiting reactant, we can divide the number of available moles of each reactant by their respective stoichiometric coefficients and find the lowest value. For N₂: \( \frac{35.7 mol~N_{2}}{1} = 35.7 \) For H₂: \( \frac{247.5 mol~H_{2}}{3} = 82.5 \) Since 35.7 is the lowest value, N₂ is the limiting reactant.
03

Calculate the maximum mass of ammonia produced

Using the stoichiometry of the balanced equation and the limiting reactant, we can find the moles of ammonia produced. In this reaction, one mole of N₂ produces two moles of NH₃. The molar mass of NH₃ is 17.03 g/mol. \( Moles~of~NH_{3} = 35.7 mol~N_{2} \times \frac{2~mol~NH_{3}}{1~mol~N_{2}} = 71.4 mol~NH_{3} \) Now convert the moles of ammonia produced into mass: \( Mass~of~NH_{3} = 71.4 mol~NH_{3} \times \frac{17.03 g~NH_{3}}{1 mol~NH_{3}} = 1.22\times 10^3 g~NH_{3} \) The maximum mass of ammonia that can be produced is \(1.22\times 10^3~g~NH_{3}\). #b. Finding the mass of the leftover starting material#
04

Calculate the moles of the leftover starting material

Since N₂ is the limiting reactant, we need to determine the moles of hydrogen that were not consumed in the reaction. We can use the stoichiometry of the balanced equation again: \( Moles~of~unused~H_{2} = 35.7 mol~N_{2} \times \frac{3~mol~H_{2}}{1~mol~N_{2}} = 107.1 mol~H_{2} \) Next, find the remaining moles of H₂: \( Remaining~H_{2} = Initial~H_{2} - Used~H_{2} = 247.5 mol~H_{2} - 107.1 mol~H_{2} = 140.4 mol~H_{2} \)
05

Convert the moles of the leftover starting material into mass

Finally, convert the moles of leftover hydrogen gas into mass: \( Mass~of~leftover~H_{2} = 140.4 mol~H_{2} \times \frac{2.02 g~H_{2}}{1 mol~H_{2}} = 283.6 g~H_{2} \) The mass of the leftover starting material, hydrogen gas, is approximately 283.6 g.

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Most popular questions from this chapter

Freon- \(12\left(\mathrm{CCl}_{2} \mathrm{~F}_{2}\right)\) is used as a refrigerant in air conditioners and as a propellant in aerosol cans. Calculate the number of molecules of Freon-12 in \(5.56 \mathrm{mg}\) of Freon-12. What is the mass of chlorine in \(5.56 \mathrm{mg}\) of Freon- \(12 ?\)

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