**Problem Statement:**If 2.60 moles of \( \text{H}_2 \) and 1.55 moles of \( \text{O}_2 \) react, how many moles of \( \text{H}_2\text{O} \) can be produced in the reaction below?\[ 2 \, \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \, \text{H}_2\text{O}(g) \]**Explanation:**In the balanced chemical equation provided, two moles of hydrogen gas (\( \text{H}_2 \)) react with one mole of oxygen gas (\( \text{O}_2 \)) to produce two moles of water (\( \text{H}_2\text{O} \)).**Steps to Solve:**1. **Determine the Limiting Reactant:** - From the equation, the ratio of \( \text{H}_2 \) to \( \text{O}_2 \) is 2:1. - Calculate the required moles of \( \text{O}_2 \) for 2.60 moles of \( \text{H}_2 \): - \( 2.60 \, \text{moles} \, \text{H}_2 \times \frac{1 \, \text{mole} \, \text{O}_2}{2 \, \text{moles} \, \text{H}_2} = 1.30 \, \text{moles} \, \text{O}_2 \) - Available \( \text{O}_2 \) is 1.55 moles, which is more than the 1.30 moles required, so \( \text{H}_2 \) is the limiting reactant.2. **Calculate Moles of \( \text{H}_2\text{O} \) Produced:** - According to the stoichiometry, 2 moles of \( \text{H}_2 \) produce 2 moles of \( \text{H}_2\text{O} \). - \( 2.60 \, \text{moles} \, \text{H}_2 \) will thus produce \( 2.60 **Question:**If you have 5 mol of \( \text{H}_2 \) and 2 mol of \( \text{N}_2 \), what is the limiting reagent in the reaction below?\[ 3 \text{H}_2 (\text{g}) + \text{N}_2 (\text{g}) \rightarrow 2 \text{NH}_3 (\text{g}) \]**Options:**- A) \( \text{H}_2 \)- B) \( \text{N}_2 \)- C) \( \text{NH}_3 \)- D) Both reactants are limiting.**Explanation:**This question tests the understanding of the concept of limiting reagents in a chemical reaction. The balanced chemical equation indicates that 3 moles of \( \text{H}_2 \) are required for every 1 mole of \( \text{N}_2 \) to produce ammonia (\( \text{NH}_3 \)). By calculating the number of moles needed and comparing it with what's available, one can determine which reactant is limiting.

Limiting reagent : The reactant which has the lowest value of (Number of moles / coefficient ) , is the limiting reagent.Here for the first reaction ; 2H2(g)+O2(g) = 2H2O (g) For H2: (Number of moles / coefficient )= 2.60/2 = 1.30 For O2 : 1.55/1=1.55 As for H2 ,the (Number of moles / coefficient ) value is minimum so, H2 is the limiting reagent. now , From 2 mole H2 produced H2O is 2 mole So ,from 2.60 mole H2 ,produced H2O =(2/2)×2.60= 2.60 mole Ans : 2.60 moles water can be produced. The reaction ; 3H2(g) + N2(g) = 2 NH3 For H2 : (Number of moles / coefficient ) is = (5/3) = 1.67 For N2: (Number of moles / coefficient ) = (2/1) = 2 So, H2 has the lowest (Number of moles / coefficient ) value. So, H2 is the limiting reagent.

Answered: 2 If 2.60 moles of H₂ and 1.55 moles of…

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2 If 2.60 moles of H₂ and 1.55 moles of O₂ react how many moles of H₂O can be produced in the reaction below? 2 H₂(g) + O₂(g) → 2 H₂O(g)

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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