For a given unbalanced reaction,MnO_2 + HCl r | vedclass

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(B) The balanced chemical equation is: $MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2$ Given mass of $MnO_2 = 100 \ g$. Moles of $MnO_2 = \frac{100 \

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$HCl$

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(B) The balanced chemical equation is: $MnO_2 + 4HCl ightarrow MnCl_2 + 2H_2O + Cl_2$ Given mass of $MnO_2 = 100 \ g$. Moles of $MnO_2 = \frac{100 \ g}{86.9 \ g/mol} = 1.15 \ mol$. Given mass of $HCl = 100 \ g$. Moles of $HCl = \frac{100 \ g}{36.5 \ g/mol} = 2.74 \ mol$. From the stoichiometry,$1 \ mol$ of $MnO_2$ reacts with $4 \ mol$ of $HCl$. Therefore,$1.15 \ mol$ of $MnO_2$ require $1.15 	imes 4 = 4.6 \ mol$ of $HCl$. Since the available moles of $HCl$ $(2.74 \ mol)$ are less than the required moles $(4.6 \ mol)$,$HCl$ is the limiting reagent.

For a given unbalanced reaction,$MnO_2 + HCl \rightarrow MnCl_2 + H_2O + Cl_2$,which is the limiting reagent if the initial amount for each reactant is $100 \ g$? [Molar masses: $MnO_2 = 86.9 \ g/mol$; $HCl = 36.5 \ g/mol$]

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