If $5\mathrm{.}0$ grams of $NaBi{O}_3$ are used to oxidize $Mn{C}_2$ into $Mr{O}_4,.$ Iour many grams of $MnC{l}_2$ Ctan be oxidized? $($Molar Mass of $NaEB{O}_3*304\frac{g}{m}ol,MnC{l}_2=126\frac{g}{m}olNaBi{O}_3+2MnC{l}_2+4H_{2}O$ Bi$((OH{)}_3+2Mn{O}_4^{-}+2C{l}^{+}+N{a}^{+}8\mathrm{.}28g6\mathrm{.}21g4\mathrm{.}1482\mathrm{.}07g$ If $5\mathrm{.}0$ grams of $NaBi{O}_3$ are used to oxidize $Mn{C}_2$ into $Mr{O}_4,.$ Iour many grams of $MnC$