The moles of methane required to produce 81 | vedclass

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

Question

(B) The balanced chemical equation for the combustion of methane is:$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$From the stoichiometry,$1 \ mol$ of $CH_4$ p

Vedclass · Accepted Answer

$225$

Vedclass · Answer

(B) The balanced chemical equation for the combustion of methane is:$CH_4 + 2O_2 ightarrow CO_2 + 2H_2O$From the stoichiometry,$1 \ mol$ of $CH_4$ produces $2 \ mol$ of $H_2O$.Given mass of water = $81 \ g$.Molar mass of $H_2O = 18 \ g/mol$.Moles of $H_2O$ produced = $\frac{81 \ g}{18 \ g/mol} = 4.5 \ mol$.Since $2 \ mol$ of $H_2O$ are produced from $1 \ mol$ of $CH_4$,the moles of $CH_4$ required = $\frac{4.5}{2} = 2.25 \ mol$.Expressing $2.25 \ mol$ in terms of $10^{-2} \ mol$:$2.25 = 225 	imes 10^{-2} \ mol$.Thus,the nearest integer is $225$.

The moles of methane required to produce $81 \ g$ of water after complete combustion is $.... \times 10^{-2} \ mol.$ [nearest integer]

Similar Questions

How many molecules of carbon dioxide are formed when $0.6 \ g$ carbon is burnt in air?

The volume of gas at $NTP$ produced by $100 \ g$ of $CaC_2$ with water is $:- CaC_2 + 2 H_2O \rightarrow Ca(OH)_2 + C_2H_{2(g)}$ (in $L$)

What volume in $mL$ of $96\% \, w/w \, H_2SO_4$ is required to prepare $2 \, L$ of $3 \, M \, H_2SO_4$ solution? (Density of $H_2SO_4$ solution $= 1.83 \, g/mL$)

What is the weight in $g$ of potassium chlorate $(KClO_3)$ required to produce the oxygen needed for the complete combustion of $84 \ L$ of $CO$ at $N.T.P$?

The molarity of $HNO_{3}$ in a sample which has density $1.4 \; g/mL$ and mass percentage of $63\%$ is . . . . . . $M$.
(Molecular Weight of $HNO_{3} = 63$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module