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Chapter 8: Problem 93

A mixture of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) is caused to react in a closed container to form \(\mathrm{NH}_{3} .\) The reaction ceases before either reactant has been totally consumed. At this stage, \(2.0 \mathrm{moles}\) each of \(\mathrm{N}_{2}, \mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are present. The moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) present originally were, respectively, (a) 4 and 4 moles (b) 3 and 5 moles (c) 3 and 4 moles (d) 4 and 5 moles

Short Answer

Expert verified
Answer: The initial moles of N₂ and H₂ present before the reaction stopped were 3 moles and 5 moles, respectively.

Step by step solution

01

Write the balanced chemical equation

First, write down the balanced chemical equation for the given reaction: $$N_{2} + 3H_{2} \rightarrow 2NH_{3}$$ This equation tells us for every mole of N₂ that reacts, 3 moles of H₂ also react, and 2 moles of NH₃ are formed.
02

Identify the mole fractions

We are given that 2.0 moles each of N₂, H₂, and NH₃ are present when the reaction stops. Let's represent the initial moles of N₂ and H₂ as x and y, respectively. Since the reaction stops, we can write the mole fractions of each substance: N₂: x - 2 H₂: y - 6 NH₃: 2 Here, x and y are the initial moles of N₂ and H₂, respectively.
03

Examine the mole ratios

Now, considering the balanced equation and the mole ratios, we can write the equation for the reaction extent: $$\frac{x - 2}{1} = \frac{y - 6}{3}$$
04

Find the possible values for x and y

We can rewrite the equation as: $$3(x - 2) = y - 6$$ Now, we need to find x and y that satisfy this equation and match one of the given options. We can check the options one by one: (a) 4 and 4 moles: The equation becomes 3(4-2)=4-6, which is false. (b) 3 and 5 moles: The equation becomes 3(3-2)=5-6, which is true. (c) 3 and 4 moles: The equation becomes 3(3-2)=4-6, which is false. (d) 4 and 5 moles: The equation becomes 3(4-2)=5-6, which is false.
05

Identify the correct answer

Based on the calculated values of x and y in step 4, only option (b) 3 and 5 moles satisfies the equation. Therefore, the moles of N₂ and H₂ present initially were 3 moles and 5 moles, respectively.

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