What volume of 0.255 M hydrochloric acid reacts completely with 0.400 g of sodium hydrogen carbonate,NaHCO₃ (84.01 g/mol)? NaHCO₃(s)+ HCl(aq)→ NaCl(aq)+ H₂O(l)+ CO₂(g)

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Quizplus · Accepted Answer

First, calculate the moles of NaHCO₃: $0.400 \, \text{g} \div 84.01 \, \text{g/mol} = 0.00476 \, \text{mol}$. The reaction is 1:1, so $0.00476 \, \text{mol}$ of HCl is needed. To find the volume of HCl: $0.00476 \, \text{mol} \div 0.255 \, \text{M} = 0.0187 \, \text{L}$, or $18.7 \, \text{mL}$.

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