Question

What volume of decinormal ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution would be needed to convert $1.19\text{g}$ of Tin to Stannic chloride in an acidic medium?

Answer 1

Step 1: Write the reactions taking place and determine the value of $n$

${\mathrm{Sn}}^0\to {\mathrm{Sn}}^{4+}+4{\mathrm{e}}^{-}$ ($n=4$)

${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}$ ($n=6$)

Step 2: Apply the law of equivalence

Equivalence of ${\mathrm{Sn}}^{4+}$ $=$Equivalence of ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}$

$$\begin{gathered} \frac{1.19}{119}\times 4=\frac{1}{10}\times V \\ V=0.01\times 4\times 10 \\ V=0.4\mathrm{L}=400\mathrm{mL} \end{gathered}$$

Hence, the volume of decinormal ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution needed to convert $1.19\mathrm{g}$ of Tin to Stannic chloride in an acidic medium is $400\mathrm{mL}$.