Question

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

2HCl(aq) + BaCO3(s) ebab9d75-6e9a-4c80-9bcd-b491b668f097.gifBaCl2(aq) + H2O(l) + CO2(g)

_____ mL

Answer 1

197mL of 0,506M HCl

Step-by-step 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 ×
`(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 ×
`(2molHCl)/(1molBaCO_(3))` = 0,0998 moles of HCl

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

0,0998 moles of HCl×
`(1L)/(0,506moles)` = 0,197 L ≡ 197mL

I hope it helps!