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What is the minimum amount (ml) of 6.00 M H2SO4 necessary to produce 5.03 g of hydrogen gas when H,SO4 reacts with excess aluminum metal?

What is the minimum amount (ml) of 6.00 M H2SO4 necessary to produce 5.03 g of hydrogen gas when H,SO4 reacts with excess aluminum metal?

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 Calculation: Sulfuric Acid (H₂SO₄) and Hydrogen Gas Production** --- **Problem Statement:** What is the minimum amount (ml) of 6.00 M H₂SO₄ necessary to produce 5.03 g of hydrogen gas when H₂SO₄ reacts with excess aluminum metal? [ ________________ ] **Detailed Explanation:** In this problem, we are asked to find the volume of 6.00 M sulfuric acid (H₂SO₄) needed to produce a specified amount of hydrogen gas (H₂) when it reacts with excess aluminum (Al). This type of problem involves stoichiometric calculations based on a balanced chemical reaction. Key steps to solve the problem: 1. **Write the balanced chemical equation for the reaction.** 2. **Use stoichiometry to relate the mass of hydrogen gas to the moles of H₂SO₄.** 3. **Calculate the volume of H₂SO₄ solution needed.** Balanced chemical equation: \[ 2 Al + 3 H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3 H_2 \] 1. **Find moles of hydrogen gas (H₂) produced:** \[ \text{Molar mass of H₂} = 2 \text{g/mol} \] \[ \text{Moles of H₂} = \frac{\text{mass of H₂}}{\text{molar mass of H₂}} = \frac{5.03 \text{g}}{2 \text{g/mol}} = 2.515 \text{mol} \] 2. **Determine moles of H₂SO₄ required:** According to the balanced equation, 3 moles of H₂ are produced by 3 moles of H₂SO₄. \[ \text{Moles of H₂SO₄} = \text{Moles of H₂} = 2.515 \text{mol} \] 3. **Calculate volume of H₂SO₄ solution:** \[ \text{Concentration of H₂SO₄} = 6.00 \text{M} \] \[ \text{Volume of H₂SO₄} = \frac{\text{Moles of H₂SO₄}}{\text{
Transcribed Image Text:**Chemical Reaction Calculation: Sulfuric Acid (H₂SO₄) and Hydrogen Gas Production** --- **Problem Statement:** What is the minimum amount (ml) of 6.00 M H₂SO₄ necessary to produce 5.03 g of hydrogen gas when H₂SO₄ reacts with excess aluminum metal? [ ________________ ] **Detailed Explanation:** In this problem, we are asked to find the volume of 6.00 M sulfuric acid (H₂SO₄) needed to produce a specified amount of hydrogen gas (H₂) when it reacts with excess aluminum (Al). This type of problem involves stoichiometric calculations based on a balanced chemical reaction. Key steps to solve the problem: 1. **Write the balanced chemical equation for the reaction.** 2. **Use stoichiometry to relate the mass of hydrogen gas to the moles of H₂SO₄.** 3. **Calculate the volume of H₂SO₄ solution needed.** Balanced chemical equation: \[ 2 Al + 3 H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3 H_2 \] 1. **Find moles of hydrogen gas (H₂) produced:** \[ \text{Molar mass of H₂} = 2 \text{g/mol} \] \[ \text{Moles of H₂} = \frac{\text{mass of H₂}}{\text{molar mass of H₂}} = \frac{5.03 \text{g}}{2 \text{g/mol}} = 2.515 \text{mol} \] 2. **Determine moles of H₂SO₄ required:** According to the balanced equation, 3 moles of H₂ are produced by 3 moles of H₂SO₄. \[ \text{Moles of H₂SO₄} = \text{Moles of H₂} = 2.515 \text{mol} \] 3. **Calculate volume of H₂SO₄ solution:** \[ \text{Concentration of H₂SO₄} = 6.00 \text{M} \] \[ \text{Volume of H₂SO₄} = \frac{\text{Moles of H₂SO₄}}{\text{
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