## Chemistry: Molecular Approach (4th Edition)

$$1.1 \space g \space of \space HCl$$
1. Write the reaction: Sodium bicarbonate: $NaHCO_3$ Hydrochloric acid: $HCl$ - This is a neutralization reaction; therefore, the acid (HCl) will donate a $H^+$ to $HC{O_3}^-$, producing $H_2CO_3$ and $NaCl$. $$HCl(aq) + NaHCO_3(aq) \longrightarrow H_2CO_3(aq) + NaCl(aq)$$ $NaHCO_3$ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 84.01 g/mol $$\frac{1 \space mole \space NaHCO_3 }{ 84.01 \space g \space NaHCO_3 } \space and \space \frac{ 84.01 \space g \space NaHCO_3 }{1 \space mole \space NaHCO_3 }$$ $HCl$ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol $$\frac{1 \space mole \space HCl }{ 36.46 \space g \space HCl } \space and \space \frac{ 36.46 \space g \space HCl }{1 \space mole \space HCl }$$ $$2.5 \space g \space NaHCO_3 \times \frac{1 \space mole \space NaHCO_3 }{ 84.01 \space g \space NaHCO_3 } \times \frac{ 1 \space mole \space HCl }{ 1 \space mole \space NaHCO_3 } \times \frac{ 36.46 \space g \space HCl }{1 \space mole \space HCl } = 1.1 \space g \space HCl$$