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Margaret [11]
3 years ago
8

How much Zn and how much HCl should be used to produce 1.5 g ZnCl2

Mathematics
1 answer:
tatuchka [14]3 years ago
4 0

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

_____

<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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