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 singl

Question

Keith_Richards [23]

2 years ago

9

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 singl

e-replacement reaction below? Show all steps of your calculation as well as the final answer. Ca + H2O → Ca(OH) 2 + H2

Chemistry

1 answer:

Nana76 [90] · Answer 1 · 2022-04-10T19:20:46+03:00

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) = <span>1.55 g/mL.
V(Ca) = m(Ca)</span>÷d(Ca) = <span>53,24g</span>÷1.55 g/mL = 34,35 mL.