How many mL of 0.506 M HClare needed to dissolve 9.85 g of BaCO3?
1 answer:
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 × = <em>0,0499 moles of BaCO₃</em>
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 × =<em> 0,0998 moles of HCl</em>
If you have a 0,506M HCl, you need to add:
0,0998 moles of HCl× = 0,197 L ≡ 197mL
I hope it helps!
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Answer:
The percentage (by mass) of KBr in the original mixture was 33.1%.
Explanation:
The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:
Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence
But the 0.042 moles came from the originals KCl and KBr moles, thus
Now it is possible to propose a system of equations:
Solving the system of equations,
0.010 moles of KBr multiplied it molar mass is
Therefore, the percentage (by mass) of KBr in the original mixture was:
%