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?

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

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Alleei Alleei · Answer 2 · 2019-10-22T09:21:20+02:00

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

Explanation :

First we have to calculate the moles of silver acetate.

[tex]\text{Moles of }CH_3COOAg=\text{Concentration of }CH_3COOAg\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }CH_3COOAg=0.500M\times 0.020L=0.01mole[/tex]

Now we have to calculate the moles of aluminum chloride.

The balanced chemical reaction will be:

[tex]AlCl_3+3CH_3COOAg\rightarrow 3AgCl+(CH_3COO)_3Al[/tex]

From the balanced reaction we conclude that,

As, 3 moles of [tex]CH_3COOAg[/tex] react with 1 mole of [tex]AlCl_3[/tex]

So, 0.01 moles of [tex]CH_3COOAg[/tex] react with [tex]\frac{0.01}{3}=3.3\times 10^{-3}[/tex] mole of [tex]AlCl_3[/tex]

Now we have to calculate the volume of aluminum chloride.

[tex]\text{Volume of }AlCl_3=\frac{\text{Moles of }AlCl_3}{\text{Concentration of }AlCl_3}\frac{3.3\times 10^{-3}mole}{0.250M}=0.0132L=13.2mL[/tex]

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