Respuesta :
[tex]\underline{\underline{\large\bf{Given:-}}}[/tex]
[tex]\red{\leadsto}\:[/tex][tex]\textsf{Amount of Methane,} [/tex][tex]\sf CH_4 = 42\:grams [/tex]
[tex]\underline{\underline{\large\bf{To Find:-}}}[/tex]
[tex]\orange{\leadsto}\:[/tex][tex]\textsf{Moles of water produced from reaction } [/tex][tex]\sf [/tex]
[tex]\\[/tex]
[tex]\underline{\underline{\large\bf{Solution:-}}}\\[/tex]
[tex]\sf Molar \:mass \:of \: CH_4 :- [/tex][tex]\sf \implies 1\times (atomic \: mass \:of \:Carbon )+ 4\times [/tex]
[tex]\sf (atomic\:mass \: of \: hydrogen)[/tex]
[tex]\sf \implies 1 \times 12+ 4\times 1 [/tex]
[tex]\sf \implies 16 grams [/tex]
[tex]\green{ \underline { \boxed{ \sf{Moles \: of \: CH_4= \frac{Amount \:of \: CH_4}{Molar \: mass \: of \:CH_4} }}}}[/tex]
[tex]\begin{gathered}\\\implies\quad \sf \frac{42}{16} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf 2.625 moles \\\end{gathered} [/tex]
The balanced reaction for combustion of methane is-
[tex]\green{ \underline { \boxed{ \sf{CH_4+2O_2 \longrightarrow CO_2+2H_2O}}}}[/tex]
By observation-
[tex]\longrightarrow[/tex]Water produced by 1 mole of methane = 2 moles
[tex]\longrightarrow[/tex]Water produced by 2.625 mole of methane = 2 × 2.625
[tex]\quad\implies[/tex] 5.250 moles
