Answer :
Final answer:
To calculate the volume of 6.50 M HCl needed to react with the zinc in the ore, we first determine the mass of Zn, then the moles of Zn, use stoichiometry to find moles of HCl, and finally convert moles of HCl to volume. The calculated volume is 7.28 mL, which does not match any of the provided options, hinting at a possible error in the question data.
Explanation:
To find out how many milliliters of 6.50 M HCl (aq) are required to react with 5.95 g of an ore containing 26.0% Zn (s) by mass, we need to use stoichiometry and the balanced equation for the reaction between zinc and hydrochloric acid:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g).
Steps to solve this problem:
- Calculate the mass of pure Zn in the ore: (5.95 g ore) × (0.260) = 1.547 g Zn.
- Determine the number of moles of Zn: moles Zn = mass Zn / molar mass Zn (65.38 g/mol) = 1.547 g / 65.38 g/mol = 0.02366 mol Zn.
- Find the number of moles of HCl needed using the stoichiometry of the reaction: 0.02366 mol Zn × (2 mol HCl / 1 mol Zn) = 0.04732 mol HCl.
- Calculate the volume of HCl solution needed: volume HCl = moles HCl / molarity HCl = 0.04732 mol / 6.50 M = 0.00728 L.
- Convert liters to milliliters: 0.00728 L × 1000 mL/L = 7.28 mL.
Thus, 7.28 mL of 6.50 M HCl (aq) are required to react with the zinc present in 5.95 g of ore containing 26.0% Zn. However, this answer is not one of the options provided, so it is possible that there was either a typographical error in the options or in the original question's data.
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