2. 112004 ? Assume volumes are auart A sample of pure KHC_(2)O_(4)cdot H_(2)C_(2)O_(4)cdot 2H_(2)O (three replaceable hydrogens)requires 46.2 mL of 0.100 M NaOH for titration.How many milliliters of 0.100 M KMnO_(4) will the same-size sample react?

To solve this problem, we need to determine the stoichiometry of the reaction between $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ and $KMnO_{4}$, and then use the titration data to find the volume of $KMnO_{4}$ required.<br /><br />Given information:<br />- The sample of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ requires 46.2 mL of 0.100 M NaOH for titration.<br /><br />Step 1: Determine the stoichiometry of the reaction between $_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ and $KMnO_{4}$.<br />The balanced chemical equation for the reaction between $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ and $KMnO_{4}$ is:<br />$5KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O + 2KMnO_{4} + 3H_{2}SO_{4} \rightarrow 5K_{2}SO_{4} + 2MnSO_{4} + 10H_{2}O + 10CO_{2}$<br /><br />Step 2: Calculate the moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ in the sample.<br />Moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ = (Volume of NaOH × Concentration of NaOH × Stoichiometry) / Volume of sample<br />Moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ = (46.2 mL × 0.100 M × 5) / 100 mL = 0.0231 mol<br /><br />Step 3: Calculate the volume of 0.100 M $KMnO_{4}$ required to react with the same-size sample.<br />Volume of $KMnO_{4}$ = (Moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ × Stoichiometry of $KMnO_{4}$) / Concentration of $KMnO_{4}$<br />Volume of $KMnO_{4}$ = (0.0231 mol × 2) / 0.100 M = 0.462 L = 462 mL<br /><br />Therefore, the same-size sample of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ will react with 462 mL of 0.100 M $KMnO_{4}$.