Question

what is the quantity of mno4 - ion reacting, in moles? 100 ml of a 0.500 m solution of fe2+ is added to 100 ml of a 0.100 m solution of mno4 -

Explanation:

Step1: Calculate moles of Fe²⁺

Use formula $n = M\times V$. Given $M = 0.500\ M$ and $V=100\ mL = 0.1\ L$. So $n_{Fe^{2 +}}=0.500\ mol/L\times0.1\ L = 0.05\ mol$.

Step2: Calculate moles of MnO₄⁻

The balanced - chemical equation for the reaction between $MnO_{4}^{-}$ and $Fe^{2+}$ is $MnO_{4}^{-}+5Fe^{2 +}+8H^{+}
ightarrow Mn^{2+}+5Fe^{3+}+4H_{2}O$. The mole - ratio of $MnO_{4}^{-}$ to $Fe^{2+}$ is $1:5$.
Given $n_{Fe^{2 +}} = 0.05\ mol$, then $n_{MnO_{4}^{-}}=\frac{n_{Fe^{2 +}}}{5}$.
$n_{MnO_{4}^{-}}=\frac{0.05\ mol}{5}=0.01\ mol$.
Also, calculate moles of $MnO_{4}^{-}$ from its own solution data. $n_{MnO_{4}^{-}\ (from\ its\ own\ solution)}=M\times V$, where $M = 0.100\ M$ and $V = 100\ mL=0.1\ L$, so $n_{MnO_{4}^{-}\ (from\ its\ own\ solution)}=0.100\ mol/L\times0.1\ L = 0.01\ mol$. Since the $MnO_{4}^{-}$ is the limiting reactant, the quantity of $MnO_{4}^{-}$ reacting is $0.01\ mol$.

Answer:

$0.01$