The percentage of Calcium carbonate in the 2.00 g piece of rock : 45%
Further explanation
Reaction
CaCO₃ + 2HCl ⇒ CaCl₂ +CO₂ + H₂O
V HCL = 36 cm³
M HCl = 0.5 mol/dm³
mol of HCl :
[tex]\tt 0.036~dm^3(36~cm^3)\times 0.5~mol/dm^3=0.018[/tex]
mol ratio CaCO₃ : HCl = 1 : 2, so mol CaCO₃ :
[tex]\tt \dfrac{1}{2}\times 0.018=0.009[/tex]
mass CaCO₃ (MW=100 g/mol) :
[tex]\tt 0.009\times 100=0.9~g[/tex]
The percentage of Calcium carbonate :
[tex]\tt \%mass=\dfrac{mass~CaCO_3}{mass~rock}\times 100\%\\\\\%mass=\dfrac{0.9}{2}\times 100\%=45\%[/tex]
