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

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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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%

idk what the question is about

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.