Question

How much oxygen will be required to burn 8G Methane completely​

Answer 1

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } SolutioN:-}}}}}$

We know that oxygen reacts with methane to give water and Carbon dioxide. The reacⁿ is :

$\boxed{\red{\bf\blue{\bullet}\:\:\underset{\blue{Methane}}{CH_4}\:+\:\underset{\blue{Oxygen}}{2O_2}\:\longrightarrow \underset{\blue{Carbon dioxide}}{CO_2}\:+\:\underset{\blue{Water}}{2H_2O}}}$

Now in this reaction ,

• $\sf Mass\:of\:CH_4\:=\:12g+4g=16g$
• $\sf Mass\:of\: Oxygen\:=\:2\times32g=64g$

So , from the above balanced reaction we can see that 16 g of methane reacts with 64g of Oxygen .

Now , use unitary method here ,

➳ $\tt 16g \:of \:CH_4\:reacts\:with\:64g\:of\:O_2$

➳ $\tt Then \:1g\:of\:CH_4\:reacts\:with\:\dfrac{64g}{16g}\:of\:O_2$

$\green{\tt\therefore 8g\:of\:CH_4\:reacts\:with\:\dfrac{\cancel{64g}\times8g}{\cancel{16g}}=\red{32g}}$

$\boxed{\purple{\bf \dag Hence\:8g\:of\: methane\:reacts\:with\:32g\:of\: Oxygen.}}$