What is the maximum amount of Cl_2 gas produc | vedclass

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(C) The balanced chemical equation is: $MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$ Molar masses: $MnO_2 = 87 \, g/mol$,$HCl = 36.5 \, g/mol$,$Cl_

Vedclass · Accepted Answer

$0.486$

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(C) The balanced chemical equation is: $MnO_2 + 4HCl ightarrow MnCl_2 + Cl_2 + 2H_2O$ Molar masses: $MnO_2 = 87 \, g/mol$,$HCl = 36.5 \, g/mol$,$Cl_2 = 71 \, g/mol$. Initial moles: $n(MnO_2) = \frac{1}{87} \approx 0.0115 \, mol$ $n(HCl) = \frac{1}{36.5} \approx 0.0274 \, mol$ According to the stoichiometry,$1 \, mol$ of $MnO_2$ requires $4 \, mol$ of $HCl$. For $0.0115 \, mol$ of $MnO_2$,we need $0.0115 	imes 4 = 0.046 \, mol$ of $HCl$. Since we have only $0.0274 \, mol$ of $HCl$,$HCl$ is the limiting reagent. From the equation,$4 \, mol$ of $HCl$ produces $1 \, mol$ of $Cl_2$. So,$0.0274 \, mol$ of $HCl$ produces $\frac{0.0274}{4} = 0.00685 \, mol$ of $Cl_2$. Mass of $Cl_2 = 0.00685 	imes 71 \approx 0.486 \, g$.

What is the maximum amount of $Cl_2$ gas produced when $1 \, g$ of $HCl$ and $1 \, g$ of $MnO_2$ are heated together (in $, g$)?

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