Question

How many mL of 0.506 M HClare needed to dissolve 9.85 g of BaCO3?

Answer 1

Answer:

197mL of 0,506M HCl

Explanation:

The reaction of HCl + BaCO₃ is:

BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.

The moles of BaCO₃ in 9,85 g are:

9,85 g of BaCO₃ × $\frac{1mol}{197,34 g}$ = 0,0499 moles of BaCO₃

As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:

0,0499 moles of BaCO₃ × $\frac{2molHCl}{1molBaCO_{3}}$ = 0,0998 moles of HCl

If you have a 0,506M HCl, you need to add:

0,0998 moles of HCl× $\frac{1L}{0,506moles}$ = 0,197 L ≡ 197mL

I hope it helps!