Question

1. how many ml of 0.715 m hcl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid)

Explanation:

Step1: Write balanced reaction

$\ce{Na2CO3 + 2HCl -> 2NaCl + H2CO3}$

Step2: Calculate moles of $\ce{Na2CO3}$

Molar mass of $\ce{Na2CO3}$ = $2×23 + 12 + 3×16 = 106$ g/mol
Moles = $\frac{1.25\ \text{g}}{106\ \text{g/mol}} ≈ 0.01179\ \text{mol}$

Step3: Find moles of HCl needed

Mole ratio HCl:$\ce{Na2CO3}$ = 2:1
Moles HCl = $2×0.01179 ≈ 0.02358\ \text{mol}$

Step4: Calculate volume of HCl

Volume (L) = $\frac{\text{moles}}{\text{molarity}} = \frac{0.02358}{0.715} ≈ 0.03298\ \text{L}$
Volume (mL) = $0.03298×1000 ≈ 33.0\ \text{mL}$

Answer:

33.0 mL