What volume of 0.115 M hydrochloric acid reacts completely with 0.125 g of sodium carbonate,Na₂CO₃ (105.99 g/mol)? Na₂CO₃(s)+ 2 HCl(aq)→ Na₂SO₄(aq)+ 2 H₂O(l)+ 2 CO₂(g)

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First, calculate the moles of Na₂CO₃: $0.125 \, \text{g} \, \text{Na}_2\text{CO}_3 \times \frac{1 \, \text{mol}}{105.99 \, \text{g}} = 0.00118 \, \text{mol} \, \text{Na}_2\text{CO}_3$. According to the reaction, 1 mole of Na₂CO₃ reacts with 2 moles of HCl. Therefore, $0.00118 \, \text{mol} \, \text{Na}_2\text{CO}_3 \times 2 = 0.00236 \, \text{mol} \, \text{HCl}$. To find the volume of HCl needed, use $V = \frac{n}{C}$, where $n = 0.00236 \, \text{mol}$ and $C = 0.115 \, \text{M}$: $V = \frac{0.00236}{0.115} = 0.0205 \, \text{L} = 20.5 \, \text{mL}$.

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