A 0.500 g sample of pyrolusite ore is treated with 50.0 mL of 0.100 M oxalic acid in dilute sulphuric acid and warmed until all the MnO2 has reacted.…

This is a back titration. The oxalic acid initially present is $50.0\times0.100=5.00\ \mathrm{mmol}$. The permanganate consumed in the final titration is $24.0\times0.0400=0.960\ \mathrm{mmol}$. Permanganate and oxalate react as $2\mathrm{MnO_4^-}+5\mathrm{H_2C_2O_4}+6\mathrm{H^+}\to2\mathrm{Mn^{2+}}+10\mathrm{CO_2}+8\mathrm{H_2O}$, so the excess oxalic acid is $\tfrac52\times0.960=2.40\ \mathrm{mmol}$. The oxalic acid consumed by the ore is therefore $5.00-2.40=2.60\ \mathrm{mmol}$. The ore reaction is $\mathrm{MnO_2+H_2C_2O_4+2H^+\to Mn^{2+}+2CO_2+2H_2O}$, a $1:1$ ratio, so $2.60\ \mathrm{mmol}$ of $\mathrm{MnO_2}$ was present. Its mass is $2.60\times10^{-3}\times86.94=0.2260\ \mathrm{g}$, and the percentage is $0.2260/0.500\times100=45.21\%$. A common error is to treat the permanganate as reacting directly with the oxalic acid in a $1:1$ ratio, or to forget the factor $5/2$; either mistake gives a wildly different oxide content, and ignoring the back-titration altogether would suggest that all of the oxalic acid was consumed by the ore.