How many milliliters of calcium, with a density of 1.55 g/mL, are needed to produce 98.5 grams of calcium hydroxide in the single-replacement reaction below?

Show all steps of your calculation as well as the final answer.

\[ \text{Ca} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2 \]

Answer is: 34,35 mL.
Reaction: Ca + 2H₂O → Ca(OH)₂ + H₂
m(Ca(OH)₂) = 98,5g.
n(Ca(OH)₂) = m÷M = 98,5g÷74g/mol = 1,33mol.
From reaction: n(Ca):n(Ca(OH)₂) = 1:1.
n(Ca) = n(Ca(OH)₂) = 1,33mol.
m(Ca) = n(Ca)·M(Ca) = 1,33mol·40g/mol = 53,24g.
d(Ca) = 1.55 g/mL.
V(Ca) = m(Ca)÷d(Ca) = 53,24g÷1.55 g/mL = 34,35 mL.

How many milliliters of calcium, with a density of 1.55 g/mL, are needed to produce 98.5 grams of calcium hydroxide in the single-replacement reaction below? Show all steps of your calculation as well as the final answer.\[ \text{Ca} + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{H}_2 \]

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