Question

If ten volume of H2 gas reacts with 5O2 gas how many volumes of water vapours would be produced?name the law

Answer 1

Answer:

$\huge\bf{\underline{\boxed{Answer:-}}}$

$\bf\red{Question:-\:How\:many\: volumes\: of\: water\: vapours\: would\: be\: produced?}$

$\bf\green{Answer:-\:10\: volumes\: of\: water\: vapours\: would\: be\: produced.}$

$\bf\red{Question:-\:Name\: of\:the\: law.}$

$\bf\green{Answer:-\:The\: Gay-\: Lussac's\:law\: of\: gaseous\: volumes.}$

By applying the Gay-Lussac's Law of Gaseous volumes to calculate the volume of water vapour produced.

$\huge\bf{\underline{\red{Reaction:-}}}$

$2H_2 (g) + O_2 (g) ----> 2H_2O (g)$

As we know that ,

When 2 volume of $H_2(g)$ react with 1 volume of $O_2(g)$ and form 2 volume of $H_2O(g)$ .

Therefore,

We have to apply the Gay-Lussac's Law of Gaseous volumes to calculate the volume of water vapour released or produced here in this reaction that given below.

$\huge\bf{\underline{\red{Reaction:-}}}$

By applying the Gay-Lussac's Law of Gaseous volumes to calculate the volume of water vapour

$10H_2 (g) + 5O_2 (g) ----> 10H_2O (g)$

When 10 volume of $H_2(g)$ react with 5 volume of $O_2(g)$ to form 10 volume of $H_2O(g)$ .