Question

KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 g. When this mixture is heated in the presence of excess Cl2, all of the KBr is converted to KCl. If the total mass of KCl present after this reaction is 3.129 g, what percentage (by mass) of the original mixture was KBr? (HINT: Be sure that you understand why the mass of the sample has decreased. It may help if you write an equation for the reaction that converted the KBr to KCl.)

Answer 1

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 (x) multiplied by it molar mass (74.5g/mol) and the moles of KBr (y) multiplied by it molar mass (119g/mol) is the total mass of the mixture:

$x.74.5g/mol + y.119g/mol = 3.595g$

Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence

$\frac{3.129g}{74.5g/mol} = 0.042 moles$

But the 0.042 moles came from the originals KCl and KBr moles, thus

$x + y = 0.042moles$

Now it is possible to propose a system of equations:

$x.74.5g/mol + y.119g/mol = 3.595g$

$x + y = 0.042moles$

Solving the system of equations,

$x=0.032moles\\y=0.010 moles$

0.010 moles of KBr multiplied it molar mass is

$0.010molesx119g/mol = 1.19g$

Therefore, the percentage (by mass) of KBr in the original mixture was:

$\frac{1.19g}{3.595g}x100% = 33.1%$%