# What volume of 0.200 M K2C2O4 is required to completely react with 30.0 mL fo 0.100 M Fe(NO3)3?

May 9, 2017

22.5 ml

#### Explanation:

$\textsf{2 F e {\left(N {O}_{3}\right)}_{3} + 3 {K}_{2} {C}_{2} {O}_{4} \rightarrow F {e}_{2} {\left({C}_{2} {O}_{4}\right)}_{3} + 6 K N {O}_{3}}$

The number of moles of $\textsf{F e {\left(N {O}_{3}\right)}_{3}}$ is given by:

$\textsf{n = c \times v = 0.100 \times \frac{30.0}{1000} = 3.00 \times {10}^{- 3}}$

From the equation we can see that the number of moles of $\textsf{{K}_{2} {C}_{2} {O}_{4}}$ must be 3/2 times this:

$\textsf{n = 3.00 \times {10}^{- 3} \times \frac{3}{2} = 4.50 \times {10}^{- 3}}$

$\textsf{n = c \times v}$

$\therefore$$\textsf{v = \frac{n}{c} = \frac{4.50 \times {10}^{- 3}}{0.200} = 22.5 \times {10}^{- 3} \textcolor{w h i t e}{x} L}$

$\textsf{v = 22.5 \textcolor{w h i t e}{x} m l}$