Question Detail
Question
What mass of ZnCl₂ can be prepared from the reaction of 3.27 g of zinc with 3.30 g of HCl? Zn + 2HCl → ZnCl₂ + H₂
Correct Answer
Option B is the correct answer.
Detailed Explanation
Molar mass Zn = 65 g/mol, HCl = 36.5 g/mol, ZnCl₂ = 136 g/mol. Moles Zn = 3.27/65 = 0.0503 mol. Moles HCl = 3.30/36.5 = 0.0904 mol. From the equation, 1 mol Zn requires 2 mol HCl. Required HCl for all Zn = 2 × 0.0503 = 0.1006 mol. Since only 0.0904 mol HCl is available, HCl is the limiting reagent. From the equation, 2 mol HCl produce 1 mol ZnCl₂. Moles ZnCl₂ produced = (1/2) × 0.0904 = 0.0452 mol. Mass ZnCl₂ = 0.0452 × 136 = 6.147 g. The closest answer is 6.17g, considering rounding differences in atomic masses (e.g., using Zn=65.4, Cl=35.5).
Hint
Calculate the moles of each reactant to find the limiting reagent, then determine the theoretical mass of the product.