# A sample of indium chloride, "InCl"_3, is known to be contaminated with sodium chloride, NaCl. The contaminated sample is then treated with excess "AgNO"_3(aq). What is the percentage by mass of "InCl"_3 in the sample?

## A sample of indium chloride, ${\text{InCl}}_{3}$, is known to be contaminated with sodium chloride, NaCl. The contaminated sample weighs 2.55 g. When the contaminated sample is treated with excess ${\text{AgNO}}_{3}$(aq), 5.16 grams of AgCl(s) are obtained. What is the percentage by mass of ${\text{InCl}}_{3}$ in the sample?

Oct 2, 2017

Purity of $I n C {l}_{3}$ is about 84.4%.

#### Explanation:

First, write down chemical equation and see what happens:

$I n C {l}_{3} + 3 A g N {O}_{3} \to I n {\left(N {O}_{3}\right)}_{3} + 3 A g C l$
$N a C l + A g N {O}_{3} \to N a N {O}_{3} + A g C l$

Atomic weights for $N , O , C l , N a , A g$ and $I n$ are $14.0 , 16.0 , 35.5 , 23.0 , 107.9$ and $114.8$, respectively.
Then calculate the molar mass(g/mol).
$I n C {l}_{3} = 221.3$, $N a C l = 58.5$ and $A g C l = 143.4$

Now let $x$ to be amount of $I n C {l}_{3}$ in mole, and y to be that of $N a C l$.

Consider the weight of sample:
$221.3 x + 58.5 y = 2.55$ ・・・(1)
Consider the weight of $A g C l$:
$143.4 \left(3 x + y\right) = 5.16$・・・(2)

Solve the simultaneous equation and you will find:
$x = 9.72 \times {10}^{- 3} \text{moles" and y=6.84xx10^(-3) "moles}$

The amount of $I n C {l}_{3}$ is $221.3 \times 9.72 \times {10}^{- 3} = 2.15$ grams, and the percentage in mass of $I n C {l}_{3}$ is 2.15/2.55xx100=84.4%