Asked by Andrea
treatment of a 0.400g sample of impure potassium chloride with an excess of AgNO3 resulted in the formation of 0.733g of AgCl. calculate the percentage of KCl in the sample. show balanced equations in support of your answer
Answers
Answered by
bridgatte
agcl:kcl
1 : 1
n of agcl=(0.733g/143.32g.mol)=5.11*10^-3mol
n of kcl=5.11*10^-3mol
m kcl=5.11*10^-3mol(74.5513g/mol)=0.381g
%kcl=(0.381/0.400)(100)=95.25%
1 : 1
n of agcl=(0.733g/143.32g.mol)=5.11*10^-3mol
n of kcl=5.11*10^-3mol
m kcl=5.11*10^-3mol(74.5513g/mol)=0.381g
%kcl=(0.381/0.400)(100)=95.25%
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