Question

Consider the reaction of aluminum chloride with silver acetate. How ml of 0.250M aluminum chloride are required to react completely with 20ml of 0.500M silver acetate?

Answer 1

Answer : The volume of aluminum chloride required are 13.2 mL.

Explanation :

First we have to calculate the moles of silver acetate.

`\text{Moles of }CH_3COOAg=\text{Concentration of }CH_3COOAg* \text{Volume of solution}`

`\text{Moles of }CH_3COOAg=0.500M* 0.020L=0.01mole`

Now we have to calculate the moles of aluminum chloride.

The balanced chemical reaction will be:

`AlCl_3+3CH_3COOAg\rightarrow 3AgCl+(CH_3COO)_3Al`

From the balanced reaction we conclude that,

As, 3 moles of
`CH_3COOAg` react with 1 mole of
`AlCl_3`

So, 0.01 moles of
`CH_3COOAg` react with
`(0.01)/(3)=3.3* 10^(-3)` mole of
`AlCl_3`

Now we have to calculate the volume of aluminum chloride.

`\text{Volume of }AlCl_3=\frac{\text{Moles of }AlCl_3}{\text{Concentration of }AlCl_3}(3.3* 10^(-3)mole)/(0.250M)=0.0132L=13.2mL`

Therefore, the volume of aluminum chloride required are 13.2 mL.

Answer 2

The balanced chemical equation would be as follows:

AlCl3(aq) + 3AgC2H3O2(aq) -> 3AgCl(s) + Al(C2H3O2)3(aq)

We are given the concentrations of the reactants. We use these values to calculate for the volume of aluminum chloride needed. We do as follows:

0.500 mol/L AgC2H3O2 (0.020 L) ( 1 mol AlCl3 / 3 mol AgC2H3O2 ) ( 1 L / 0.250 AlCl3 ) = 0.0133 L or 13.3 mL of AlCl3 solution is needed