# Question #fe05d

Apr 26, 2016

$17.7 \text{ml}$

#### Explanation:

$N {a}_{2} S {O}_{4 \left(a q\right)} + B a C {l}_{2 \left(a q\right)} \rightarrow B a S {O}_{4 \left(s\right)} + 2 N a C {l}_{\left(a q\right)}$

A white precipitate of insoluble barium sulfate is formed.

From the equation you can see that the reactants are in a 1:1 molar ratio.

Concentration = no. moles/volume of solution.

$c = \frac{n}{v}$

$\therefore n = c \times v$

So we can work out the number of moles of barium chloride:

${n}_{B a C {l}_{2}} = 0.218 \times \frac{26.8}{1000} = 0.00584$

The equation tells us that the number of moles of sodium sulfate must be the same:

${n}_{N {a}_{2} S {O}_{4}} = 0.00584$

Since $c = \frac{n}{v}$ we can write:

$\frac{0.00584}{v} = 0.33$

$\therefore v = \frac{0.00584}{0.33} = 0.0177 \text{L}$

$v = 17.7 \text{ml}$