For the reaction 2Al + 3MnO xrightarrow{Delta | vedclass

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(D) The balanced chemical equation is: $2Al + 3MnO \rightarrow Al_2O_3 + 3Mn$. Molar masses: $Al = 27 \ g/mol$,$MnO = 55 + 16 = 71 \ g/mol$. Given mol

Vedclass · Accepted Answer

$159 \ g$ of $MnO$ is required

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(D) The balanced chemical equation is: $2Al + 3MnO ightarrow Al_2O_3 + 3Mn$. Molar masses: $Al = 27 \ g/mol$,$MnO = 55 + 16 = 71 \ g/mol$. Given moles: $n(Al) = 108/27 = 4 \ mol$,$n(MnO) = 213/71 = 3 \ mol$. According to the stoichiometry,$2 \ mol$ of $Al$ reacts with $3 \ mol$ of $MnO$. For $3 \ mol$ of $MnO$,the required $Al$ is $(2/3) 	imes 3 = 2 \ mol$. Since we have $4 \ mol$ of $Al$,$Al$ is in excess and $MnO$ is the limiting reagent. Amount of $Al$ required = $2 \ mol 	imes 27 \ g/mol = 54 \ g$. Amount of $MnO$ required for $4 \ mol$ of $Al$ = $(3/2) 	imes 4 = 6 \ mol = 6 	imes 71 = 426 \ g$. Thus,option $A$ is incorrect because $Al$ is in excess,but the statement says $Al$ is in excess (which is true),wait: Let's re-evaluate: $Al$ is in excess ($4 \ mol$ provided,$2 \ mol$ needed). $MnO$ is the limiting reagent. Option $A$: $Al$ is in excess (True). Option $B$: $MnO$ is limiting reagent (True). Option $C$: $54 \ g$ of $Al$ is required (True,for $3 \ mol$ of $MnO$). Option $D$: $159 \ g$ of $MnO$ is required (False,$426 \ g$ is required for $4 \ mol$ of $Al$). Therefore,the incorrect statement is $D$.

For the reaction $2Al + 3MnO \xrightarrow{\Delta} Al_2O_3 + 3Mn$; If $108 \ g$ of $Al$ and $213 \ g$ of $MnO$ are heated,then which of the following is incorrect?

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