108 g Al and 213 g MnO were heated to ini | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. Calculate moles of reactants:$n_{Al} = \frac{108 \ g}{27 \ g/mol} = 4 \ mol$$n_{MnO} = \frac{213 \ g}{71 \ g/mol} = 3 \ mol$$2$. Determine th

Vedclass · Accepted Answer

$102 \ g \ Al_2O_3$ is formed

Vedclass · Answer

(B) $1$. Calculate moles of reactants:$n_{Al} = \frac{108 \ g}{27 \ g/mol} = 4 \ mol$$n_{MnO} = \frac{213 \ g}{71 \ g/mol} = 3 \ mol$$2$. Determine the limiting reagent by comparing $\frac{n}{	ext{stoichiometric coefficient}}$:For $Al$: $\frac{4}{2} = 2$For $MnO$: $\frac{3}{3} = 1$Since $1 $3$. Calculate product formation based on $MnO$:From the reaction,$3 \ mol \ MnO$ produces $1 \ mol \ Al_2O_3$.Therefore,$3 \ mol \ MnO$ produces $1 \ mol \ Al_2O_3$.Mass of $Al_2O_3 = 1 \ mol 	imes (2 	imes 27 + 3 	imes 16) \ g/mol = 1 	imes 102 \ g = 102 \ g$.Thus,option $B$ is correct.

$108 \ g \ Al$ and $213 \ g \ MnO$ were heated to initiate the reaction:
$2 Al + 3 MnO \to Al_2O_3 + 3 Mn$
Which of the following is correct? (Atomic weights: $Mn = 55$,$Al = 27$)

Similar Questions

What is a limiting reagent?

$56 \ g$ of $CaO$ has been mixed with $63 \ g$ of $HNO_3$. The amount of $Ca(NO_3)_2$ formed is: (in $g$)

In the reaction $2H_2 + O_2 \to 2H_2O$,if $6 \ g$ of $H_2$ combines with $64 \ g$ of $O_2$,find the mass of the excess reagent left in $g$.

For the reaction,$2Fe(NO_3)_3 + 3Na_2CO_3 \to Fe_2(CO_3)_3 + 6NaNO_3$. Initially,if $2.5 \ mol$ of $Fe(NO_3)_3$ and $3.6 \ mol$ of $Na_2CO_3$ are taken. If $6.3 \ mol$ of $NaNO_3$ is obtained,then the $\%$ yield of the given reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$108 \ g \ Al$ and $213 \ g \ MnO$ were heated to initiate the reaction:$2 Al + 3 MnO \to Al_2O_3 + 3 Mn$Which of the following is correct? (Atomic weights: $Mn = 55$,$Al = 27$)