Mass of methane required to produce 22 g of | vedclass

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

Question

(C) The balanced chemical equation for the combustion of methane is:$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$Calculate the molar mass of $CO_2$:$M(CO_2)

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The balanced chemical equation for the combustion of methane is:$CH_4 + 2O_2 ightarrow CO_2 + 2H_2O$Calculate the molar mass of $CO_2$:$M(CO_2) = 12.0 + 2 	imes 16.0 = 44.0 \ g \ mol^{-1}$Moles of $CO_2$ produced = $\frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{22 \ g}{44 \ g \ mol^{-1}} = 0.5 \ mol$From the stoichiometry of the reaction,$1 \ mol$ of $CH_4$ produces $1 \ mol$ of $CO_2$.Therefore,$0.5 \ mol$ of $CO_2$ requires $0.5 \ mol$ of $CH_4$.Calculate the molar mass of $CH_4$:$M(CH_4) = 12.0 + 4 	imes 1.0 = 16.0 \ g \ mol^{-1}$Mass of $CH_4$ required = $	ext{Moles} 	imes 	ext{Molar mass} = 0.5 \ mol 	imes 16.0 \ g \ mol^{-1} = 8 \ g$.

Mass of methane required to produce $22 \ g$ of $CO_2$ after complete combustion is . . . . . . $g.$ (Given Molar mass in $g \ mol^{-1}$: $C=12.0, H=1.0, O=16.0$)

Similar Questions

The volume (in $mL$) of $0.1\, N$ $NaOH$ required to neutralize $10\, mL$ of $0.1\, N$ phosphinic acid is........

Commercial concentrated sulfuric acid is $95\% $ $H_2SO_4$ by weight. If the density of this commercial acid is $1.834 \ g \ cm^{-3}$,the molarity of this solution is ....... $M$.

The volume of $0.02 \, M$ aqueous $HBr$ required to neutralize $10.0 \, mL$ of $0.01 \, M$ aqueous $Ba(OH)_2$ is $....... \, mL$ (Assume complete neutralization).

The hydrated salt $Na_2SO_4 \cdot nH_2O$ undergoes $55.9 \%$ loss in weight on heating and becomes anhydrous. The value of $n$ will be?

For the reaction,$A \to nB$. At the point of intersection of the two curves in the given graph,the concentration of $B$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module