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Chapter 3: Problem 103

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g)\). The reaction ceases before either reactant has been totally consumed. At this stage $3.0 \mathrm{~mol} \mathrm{~N}_{2}, 3.0 \mathrm{~mol} \mathrm{H}_{2}\(, and \)3.0 \mathrm{~mol} \mathrm{NH}_{3}\( are present. How many moles of \)\mathrm{N}_{2}$ and \(\mathrm{H}_{2}\) were present originally?

Short Answer

Expert verified
There were initially \(4.5 \mathrm{~moles}\) of \(\mathrm{N}_{2}\) and \(7.5 \mathrm{~moles}\) of \(\mathrm{H}_{2}\) present in the container.

Step by step solution

01

Write the balanced chemical equation

Write down the balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to form ammonia gas: N2(g) + 3H2(g) → 2NH3(g) We can see the stoichiometry of the reaction is 1:3:2 for N2, H2, and NH3 respectively.
02

Identify the moles of N2 and H2 consumed by the reaction

According to the stoichiometry of the reaction, for every 2 moles of NH3 produced, 1 mole of N2 and 3 moles of H2 are consumed. We are given that 3 moles of NH3 have been formed, which means: \(Moles \quad of \quad \mathrm{N}_{2} \quad consumed = \frac{1}{2} × Moles \quad of \quad \mathrm{NH}_{3} = \frac{1}{2} × 3 = 1.5 \quad moles\) \(Moles \quad of \quad \mathrm{H}_{2} \quad consumed = \frac{3}{2} × Moles \quad of \quad \mathrm{NH}_{3} = \frac{3}{2} × 3 = 4.5 \quad moles\)
03

Determine the initial moles of N2 and H2

Now that we know the moles of N2 and H2 consumed in the reaction, we can determine the initial moles by adding them to the moles present at the equilibrium state. Moles of N2 initially: \(3.0 \, moles \, (equilibrium) + 1.5 \, moles \, (consumed) = 4.5 \, moles\) Moles of H2 initially: \(3.0 \, moles \, (equilibrium) + 4.5 \, moles \, (consumed) = 7.5 \, moles\) So, there were 4.5 moles of N2 and 7.5 moles of H2 present initially.

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Most popular questions from this chapter

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(S_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with \(30.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\), how many grams of \(S_{8}\) would be produced, assuming \(98 \%\) yield?

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