A 20.00 g mixture of magnesium and zinc metal reacting with excess hydrochloric acid produced 1.242 g of hydrogen gas.

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A 20.00 g mixture of magnesium and zinc metal reacting with excess hydrochloric acid produced 1.242 g of hydrogen gas. What is the percent zinc in the original mixture

The correct Answer and Explanation is:

To solve this problem, we need to determine the percent composition of zinc in a mixture of magnesium (Mg) and zinc (Zn) that reacts with hydrochloric acid (HCl) to produce hydrogen gas (H₂).

Step-by-step Approach:

  1. Reactions Involved:
    • Magnesium reacts with hydrochloric acid: Mg+2HCl→MgCl2+H2\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2Mg+2HCl→MgCl2​+H2​
    • Zinc reacts similarly: Zn+2HCl→ZnCl2+H2\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2Zn+2HCl→ZnCl2​+H2​
    Both reactions produce hydrogen gas, but the amounts of gas depend on the masses of the metals involved.
  2. Mass of Hydrogen Gas: The reaction produced 1.242 g of hydrogen gas. Using the molar mass of hydrogen gas (H₂), which is 2.016 g/mol, we calculate the moles of hydrogen gas produced:Moles of H2=1.242 g2.016 g/mol=0.616 mol\text{Moles of H}_2 = \frac{1.242 \text{ g}}{2.016 \text{ g/mol}} = 0.616 \text{ mol}Moles of H2​=2.016 g/mol1.242 g​=0.616 mol
  3. Stoichiometric Relationships: According to the reactions, 1 mole of magnesium or 1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the total moles of hydrogen gas is equivalent to the total moles of magnesium and zinc involved in the reaction:nMg+nZn=0.616 moln_{\text{Mg}} + n_{\text{Zn}} = 0.616 \text{ mol}nMg​+nZn​=0.616 mol
  4. Mass Relationships: The molar mass of magnesium is 24.305 g/mol, and the molar mass of zinc is 65.38 g/mol. The mass of the metals in the mixture can be expressed as:Mass of Mg=nMg×24.305 g/mol\text{Mass of Mg} = n_{\text{Mg}} \times 24.305 \text{ g/mol}Mass of Mg=nMg​×24.305 g/mol Mass of Zn=nZn×65.38 g/mol\text{Mass of Zn} = n_{\text{Zn}} \times 65.38 \text{ g/mol}Mass of Zn=nZn​×65.38 g/molThe total mass of the metals is given as 20.00 g:nMg×24.305+nZn×65.38=20.00 gn_{\text{Mg}} \times 24.305 + n_{\text{Zn}} \times 65.38 = 20.00 \text{ g}nMg​×24.305+nZn​×65.38=20.00 g
  5. Solving the System of Equations: We have two equations:nMg+nZn=0.616 moln_{\text{Mg}} + n_{\text{Zn}} = 0.616 \text{ mol}nMg​+nZn​=0.616 mol nMg×24.305+nZn×65.38=20.00 gn_{\text{Mg}} \times 24.305 + n_{\text{Zn}} \times 65.38 = 20.00 \text{ g}nMg​×24.305+nZn​×65.38=20.00 gSolving these simultaneously gives nZn≈0.243 moln_{\text{Zn}} \approx 0.243 \text{ mol}nZn​≈0.243 mol.
  6. Mass of Zinc:Mass of Zn=0.243 mol×65.38 g/mol=15.89 g\text{Mass of Zn} = 0.243 \text{ mol} \times 65.38 \text{ g/mol} = 15.89 \text{ g}Mass of Zn=0.243 mol×65.38 g/mol=15.89 g
  7. Percent Zinc:Percent Zn=15.89 g20.00 g×100=79.45%\text{Percent Zn} = \frac{15.89 \text{ g}}{20.00 \text{ g}} \times 100 = 79.45\%Percent Zn=20.00 g15.89 g​×100=79.45%

Thus, the percent zinc in the original mixture is approximately 79.45%.

Explanation:

In this problem, we calculated the percent zinc by analyzing the amount of hydrogen gas produced in a reaction between a metal mixture and hydrochloric acid. By determining the number of moles of hydrogen gas and using the stoichiometric relationships of the reactions, we solved for the moles of each metal. The total mass of the metals allowed us to solve the system of equations and find the mass and percent composition of zinc in the mixture.

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