Chemistry: Molecular Approach (4th Edition)

5.2 g of $CaCO_3$
1. Write the reaction: Hydrochloric acid: $HCl$ Calcium carbonate: $CaCO_3$ This is an acid-base reaction, where $HCl$ is the acid, and the carbonate ion ${CO_3}^{2-}$ is the base. Since $C{O_3}^{2-}$ accepts 2 protons $(H^+)$, there must be 2 HCl moles for each carbonate ion mole. $$2 \space HCl(aq) + CaCO_3(s) \longrightarrow$$ The 2 $H^+$ will react with $C{O_3}^{2-}$ to produce $H_2CO_3$, and the 2 $Cl^-$ will react with the $Ca^{2+}$, producing $CaCl_2$ $$2 \space HCl(aq) + CaCO_3(s) \longrightarrow H_2CO_3(aq) + CaCl_2(aq)$$ 2. Calculate the amount of $CaCO_3$: $HCl$ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol $$\frac{1 \space mole \space HCl }{ 36.46 \space g \space HCl } \space and \space \frac{ 36.46 \space g \space HCl }{1 \space mole \space HCl }$$ $CaCO_3$ : ( 40.08 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 100.09 g/mol $$\frac{1 \space mole \space CaCO_3 }{ 100.09 \space g \space CaCO_3 } \space and \space \frac{ 100.09 \space g \space CaCO_3 }{1 \space mole \space CaCO_3 }$$ $$3.8 \space g \space HCl \times \frac{1 \space mole \space HCl }{ 36.46 \space g \space HCl } \times \frac{ 1 \space mole \space CaCO_3 }{ 2 \space moles \space HCl } \times \frac{ 100.09 \space g \space CaCO_3 }{1 \space mole \space CaCO_3 } = 5.2 \space g \space CaCO_3$$