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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

User Snmaddula
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1 Answer

3 votes

Answer:

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!

User Atlaste
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