Question

How many moles of water are produced from the combustion of 42,0

grams of methane, CH?
CH4+O2 =
CO2 + H2O

Answer 1

`\underline{\underline{\large\bf{Given:-}}}`

`\red{\leadsto}\:`
`\textsf{Amount of Methane,}`
`\sf CH_4 = 42\:grams`

`\underline{\underline{\large\bf{To Find:-}}}`

`\orange{\leadsto}\:`
`\textsf{Moles of water produced from reaction }`
`\sf`

`\\`

`\underline{\underline{\large\bf{Solution:-}}}\\`

`\sf Molar \:mass \:of \: CH_4 :-`
`\sf \implies 1* (atomic \: mass \:of \:Carbon )+ 4*`

`\sf (atomic\:mass \: of \: hydrogen)`

`\sf \implies 1 * 12+ 4* 1`

`\sf \implies 16 grams`

`\green{ \underline { \boxed{ \sf{Moles \: of \: CH_4= (Amount \:of \: CH_4)/(Molar \: mass \: of \:CH_4) }}}}`

`\begin{gathered}\\\implies\quad \sf (42)/(16) \\\end{gathered}`

`\begin{gathered}\\\implies\quad \sf 2.625 moles \\\end{gathered}`

The balanced reaction for combustion of methane is-

`\green{ \underline { \boxed{ \sf{CH_4+2O_2 \longrightarrow CO_2+2H_2O}}}}`

By observation-

`\longrightarrow`Water produced by 1 mole of methane = 2 moles

`\longrightarrow`Water produced by 2.625 mole of methane = 2 × 2.625

`\quad\implies` 5.250 moles