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y moles of NH, - 3

y moles of NH, - 3

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Chemical Reaction Problem **Question:** How many moles of NH₃ can be produced from 2.71 moles of nitrogen in the following reaction: \[ \text{N}_2 (g) + 3 \text{H}_2 (g) \rightarrow 2 \text{NH}_3 (g) \] ### Explanation: This problem involves stoichiometry, which is the calculation of reactants and products in chemical reactions. According to the balanced chemical equation, one mole of nitrogen (\(\text{N}_2\)) reacts with three moles of hydrogen (\(\text{H}_2\)) to produce two moles of ammonia (\(\text{NH}_3\)). **Steps to Solve:** 1. **Identify the Mole Ratio:** - The mole ratio between \(\text{N}_2\) and \(\text{NH}_3\) is 1:2 according to the equation. 2. **Calculate Moles of \(\text{NH}_3\):** - For every 1 mole of \(\text{N}_2\), 2 moles of \(\text{NH}_3\) are produced. - Therefore, if 2.71 moles of \(\text{N}_2\) are used, the moles of \(\text{NH}_3\) produced will be \(2.71 \times 2 = 5.42\) moles. Thus, 5.42 moles of \(\text{NH}_3\) can be produced from 2.71 moles of nitrogen.
Transcribed Image Text:### Chemical Reaction Problem **Question:** How many moles of NH₃ can be produced from 2.71 moles of nitrogen in the following reaction: \[ \text{N}_2 (g) + 3 \text{H}_2 (g) \rightarrow 2 \text{NH}_3 (g) \] ### Explanation: This problem involves stoichiometry, which is the calculation of reactants and products in chemical reactions. According to the balanced chemical equation, one mole of nitrogen (\(\text{N}_2\)) reacts with three moles of hydrogen (\(\text{H}_2\)) to produce two moles of ammonia (\(\text{NH}_3\)). **Steps to Solve:** 1. **Identify the Mole Ratio:** - The mole ratio between \(\text{N}_2\) and \(\text{NH}_3\) is 1:2 according to the equation. 2. **Calculate Moles of \(\text{NH}_3\):** - For every 1 mole of \(\text{N}_2\), 2 moles of \(\text{NH}_3\) are produced. - Therefore, if 2.71 moles of \(\text{N}_2\) are used, the moles of \(\text{NH}_3\) produced will be \(2.71 \times 2 = 5.42\) moles. Thus, 5.42 moles of \(\text{NH}_3\) can be produced from 2.71 moles of nitrogen.
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