Question

A 15.00g solid mixture containing Ca(OH)2, among other non-basic components, was neutralized with 0.2000g of HCl. What was the mass percentage of Ca(OH)2 in the mixture? Round your answer to four significant figures

Answer 1

Answer:

1.373 wt% Ca(OH)₂

Explanation:

Sample mix = 15.0g

Ca(OH)₂(aq) + 2HCl(aq) => CaCl₂(aq) + 2H₂O(l)

moles HCl = 0.2000g / 36 g·mol⁻¹ 0.0056 mol

moles Ca(OH)₂ = 1/2(moles HCl) = 1/2(0.0056 mol) = 0.0028 mol

mass Ca(OH)₂ = 0.0028 mol ( 74 g/mol ) = 0.206 g

mass % Ca(OH)₂ = (0.206/15.0)100% = 1.373 wt%

Answer 2

The mass percentage of Ca(OH)2  is 1.333 %.

The equation of the reaction is;

Ca(OH)2(s) + 2HCl(aq) -------> CaCl2(aq) + 2H2O(l)

Number of moles of HCl = 0.2000g/36.5 g/mol = 0.00548 moles

Given that 1 mole of Ca(OH)2 reacts with 2 moles of HCl

x moles of Ca(OH)2 reacts with 0.00548 moles of HCl

x = 1 mole × 0.00548 moles/2 moles

x = 0.00274 moles

Hence, mass of Ca(OH)2 reacted = 0.00274 moles  × 74 g/mol = 0.2 g

Hence;

Mass percentage of Ca(OH)2  = 0.2 g/15.00g   ×  100/1

Mass percentage of Ca(OH)2  = 1.333 %

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