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Answer 1 · 2016-04-26T13:22:51

$17.7 ml$ $17.7"ml"$

$N a_{2} S O_{4 (a q)} + B a C l_{2 (a q)} \to B a S O_{4 (s)} + 2 N a C l_{(a q)}$ $Na_2SO_(4(aq))+BaCl_(2(aq))rarrBaSO_(4(s))+2NaCl_((aq))$

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.

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

$:.n=cxxv$

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

$n_(BaCl_2)=0.218xx26.8/1000=0.00584$

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

$n_(Na_2SO_4)=0.00584$

Since $c=n/v$ we can write:

$0.00584/v=0.33$

$:.v=0.00584/0.33=0.0177"L"$

$v=17.7"ml"$