First, we have to calculate the amount in moles of "BrO_3^-" reacted:
"n(BrO_3^-)=C(BrO_3^-)*V(BrO_3^-)=0.118M*0.02750L=3.245*10^{-3}mol"
According to the given equation, the amount of the iron(II) cations required is six time higher then that of the "BrO_3^-" anions:
"n(Fe^{2+})=6*n(BrO_3^-)=6*3.245*10^{-3}mol=0.01947mol"
Now, it the right time to calculate the molar concentration of the iron(II) cations in the starting mixture:
"C(Fe^{2+})=\\frac{n(Fe^{2+})}{V(Fe^{2+})}=\\frac{0.01947mol}{0.01300L}=1.498M"
Thus, it was 1.498 M water solution of some iron(II) salt.
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