2. ? Assume volumes are 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 $KMnO_{4}$.<br /><br />First, let's write the balanced chemical equation for the reaction between $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ and $KMnO_{4}$:<br /><br />$5KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O + 2KMnO_{4} + 3H_{2}SO_{4} \rightarrow 5K_{2}SO_{4} + 2MnSO_{4} + 10CO_{2} + 8H_{2}O$<br /><br />From the balanced equation, we can see that 5 moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ react with 2 moles of $KMnO_{4}$.<br /><br />Now, let's calculate the number of moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ in the sample:<br /><br />Moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ = (Volume of NaOH solution * Concentration of NaOH solution) / (Molar mass of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$)<br /><br />Moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ = (46.2 mL * 0.100 M) / (204.22 g/mol) = 0.0225 moles<br /><br />Since 5 moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ react with 2 moles of $KMnO_{4}$, we can set up a proportion to find the number of moles of $KMnO_{4}$ needed:<br /><br />5 moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ : 2 moles of $KMnO_{4}$ = 0.0225 moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ : x moles of $KMnO_{4}$<br /><br />Solving for x, we get:<br /><br />x = (0.0225 moles * 2 moles) / 5 moles = 0.009 moles of $KMnO_{4}$<br /><br />Now, let's calculate the volume of 0.100 M $KMnO_{4}$ solution needed to react with the sample:<br /><br />Volume of $KMnO_{4}$ solution = Moles of $KMnO_{4}$ / Concentration of $KMnO_{4}$ solution<br /><br />Volume of $KMnO_{4}$ solution = 0.009 moles / 0.100 M = 0.09 L or 90 mL<br /><br />Therefore, the same-size sample will react with 90 mL of 0.100 M $KMnO_{4}$ solution.