Question 4b784

Jan 17, 2017

Let the mass of $M g$ in $36 g$ mixture be $x g$ and the mass of Ca be $\left(36 - x\right) g$
Considering atomic masses as follows

$M g \to 24 g \text{/mol}$

$C a \to 40 g \text{/mol}$

$B r \to 80 g \text{/mol}$

Balanced equations of ractions are
$M g + B {r}_{2} = M g B {r}_{2.} \ldots . . \left[1\right]$

$C a + B {r}_{2} = C a B {r}_{2.} \ldots \ldots . \left[2\right]$

As per equation 

$24 g$ Mg forms $184 g$ $M g B {r}_{2}$

$x g \text{ }$ Mg forms $\frac{184 x}{24} g \text{ } M g B {r}_{2}$

As per equation 

$40 g$ Ca forms $200 g$ $C a B {r}_{2}$

$\left(36 - x\right) g \text{ }$ Ca forms $\frac{\left(36 - x\right) 200}{20} g \text{ } C a B {r}_{2}$

Hence by the given condition of the problem we can write

$\frac{184}{24} x + \left(36 - x\right) \frac{200}{40} = 212$

$\implies \frac{23}{3} x + \left(180 - 5 x\right) = 212$

$\implies \frac{23}{3} x - 5 x = 212 - 180 = 32$

$\implies \frac{23 x - 15 x}{3} = 32$

$\implies \frac{8 x}{3} = 32$

$\implies x = 12$

So the mass of Mg is $12 g$

and the mass of Ca is $\left(36 - 12\right) g = 24 g$

% "of Mg"=12/36xx100%=33 1/3%

% "of Ca"=24/36xx100%=66 2/3%#