Blackboard chalk is a mixture of limestone and gypsum. A piece of blackboard chalk weighing \(5.42 g\) was immersed in \(HCl\). The \(CO _{2}\) produced weighed \(1.11 g\). What is the percentage of limestone in the calk?
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
\(CaCO _{3( e )}+2 HCl _{(\text {an) }} \rightarrow CaCl _{2(\infty)}+ H _{2} O _{(n+1}+ CO _{2(\text { a) }}\)
Molar mass of \(CaCO _{3}=100 gmol ^{-1}\)
Molar mass of \(CO _{2}=44 gmol ^{-1}\)ccording to the equation, 1 mole of \(CaCO _{3}\) produces 1 mole of \(CO _{2}\).
Mole of \(CO _{2}=\frac{1.11 g }{44 gmol ^{-1}}\)
\(\Rightarrow 0.02523 mol\)
Therefore, mole of \(CaCO\),
\(\Rightarrow 0.02523 mol\)
Mass of \(CaCO _{3}\) present
\(=(0.02523 \times 100) g =2.523 g\)
\(\%\) of \(CaCO _{3}=\frac{2.523 g }{5.42 g } \times 100 \%\)
\(=46.55 \%\)
\(CaCO _{3( e )}+2 HCl _{(\text {an) }} \rightarrow CaCl _{2(\infty)}+ H _{2} O _{(n+1}+ CO _{2(\text { a) }}\)
Molar mass of \(CaCO _{3}=100 gmol ^{-1}\)
Molar mass of \(CO _{2}=44 gmol ^{-1}\)ccording to the equation, 1 mole of \(CaCO _{3}\) produces 1 mole of \(CO _{2}\).
Mole of \(CO _{2}=\frac{1.11 g }{44 gmol ^{-1}}\)
\(\Rightarrow 0.02523 mol\)
Therefore, mole of \(CaCO\),
\(\Rightarrow 0.02523 mol\)
Mass of \(CaCO _{3}\) present
\(=(0.02523 \times 100) g =2.523 g\)
\(\%\) of \(CaCO _{3}=\frac{2.523 g }{5.42 g } \times 100 \%\)
\(=46.55 \%\)