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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Answer:
The mean, μ is 2.5063 and
The standard deviation, σ = 2.0499 × 10⁻²
Step-by-step explanation:
Here we have
and
10% have length < 2.48 in while
5% have length > 2.54 in
and
From the z score table
Critical z at 10% to the left = -1.282
Critical z at 5% to the right = 1.645
Therefore, the equations become
-1.282σ = 2.48 - μ → μ -1.282σ = 2.48
1.645σ = 2.54 - μ → μ + 1.645σ = 2.54
Solving the above equations gives
The mean, μ = 2.5063 and
The standard deviation, σ = 2.0499 × 10⁻².