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Assume the Ca piece was pure (not partially oxidized) and weighed 4.5645 g. Assuming the reaction goes to completion, how many grams of Ca(OH)₂ should you have made? Use the molecular weight of Ca(OH)₂ to 5 sig figs.

User Mike James
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Final answer:

To find the mass of Ca(OH)2 produced from 4.5645 g of pure Ca, calculate the number of moles of Ca and multiply by the molar mass of Ca(OH)2 to get 8.438 g of Ca(OH)2.

Step-by-step explanation:

To calculate the expected mass of Ca(OH)₂ produced from the reaction of 4.5645 g of pure Ca, stoichiometry of the reaction and molar masses of the substances involved must be used. First, the molar mass of Ca(OH)₂ must be calculated: 40.08 g/mol for Ca, 2 imes1.008 g/mol for H, and 2 imes16.00 g/mol for O, giving a total of 74.096 g/mol to five significant figures for Ca(OH)₂.

Next, one should note the balanced chemical equation for the reaction, which is typically:
Ca + 2H₂O → Ca(OH)₂ + H₂.

According to this equation, one mole of Ca will produce one mole of Ca(OH)₂. Therefore, the number of moles of Ca used, which is the mass of Ca divided by its molar mass (40.08 g/mol), needs to be calculated. Finally, this mole quantity is multiplied by the molar mass of Ca(OH)₂ to yield the mass of the product formed.

Specifically, for 4.5645 g of Ca:
4.5645 g Ca × (1 mol Ca / 40.08 g) = 0.1139 mol Ca
0.1139 mol Ca(OH)₂ × (74.096 g/mol) = 8.438 g of Ca(OH)₂

User Tyleha
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