Question

称取铁矿石200.0mg,滴定时用去c(1/6K2Cr2O7)=0.05mol/L的重铬酸钾标准溶液35.82mL计算矿

Answer 1

n(Fe) =n(1/6K2Cr2O7)=0.05mol/L *35.82 mL=1.791mmol

计算矿石中铁的百分含量=(1.791mmol*55.85g/mol) /200mg *100%=50%

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也可这样做

c(1/6K2Cr2O7)=6c( K2Cr2O7)=0.05mol/L

c( K2Cr2O7)=0.05mol/L/6

6Fe -----6Fe2+ --------------K2Cr2O7

6 1

200mg *x /55.85mg/mol 0.05/6 mol/L *35.82mL

x=0.5