How much Zn and how much HCl should be used to produce 1.5 g ZnCl2
1 answer:
Answer:
- 0.71963 g of Zn
- 0.80256 g of HCl
Step-by-step explanation:
The required mass of reactants can be computed from the number of moles of product.
moles of ZnCl₂ = (1.5 g)/(136.282 g) ≈ 0.0110066 moles
Then the mass of Zn required is ...
mass of Zn = (0.0110066 mol)(65.382 g/mol) = 0.71963 g
Two moles of HCl are used in the reaction with each mole of Zn, so the mass of HCl required is ...
mass of HCl = 2(0.0110066 mol)(36.458 g/mol) = 0.80256 g
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<em>Comment</em>
The discrepancy in the last two decimal places of the weights above versus the calculator output below comes from rounding of moles of Zn to 4 significant digits, instead of 5 or more. The number of moles of ZnCl₂ is closer to 0.0110066.
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