Question

Methane reacts with steam to form ${{H}_{2}}$ and CO as shown. What volume of ${{H}_{2}}$ can be obtained from 100 $c{{m}^{3}}$ of methane at STP?
$C{{H}_{4}}+{{H}_{2}}O\to CO+{{H}_{2}}$
(a)- 100 $c{{m}^{3}}$
(b)- 150 $c{{m}^{3}}$
(c)- 300 $c{{m}^{3}}$
(d)- 200 $c{{m}^{3}}$

Answer 1

Hint:To find the volume of the compound you have to first balance the reaction correctly because it will help to find the correct number of moles. We can use a formula to solve this question is:
$\dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{{{V}_{1}}}{{{V}_{2}}}$
Where n and V are the numbers of moles are volume of the compounds respectively.

Complete step-by-step answer:We can solve this question on the basis of "Avogadro's law" which states that "Equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of moles." So we can say that volume of the gas is directly proportional to the number of moles. We can write:
$V\propto n$
$\dfrac{V}{n}=k$
To find the volume of the compound you have to first balance the reaction correctly because it will help to find the correct number of moles. The given reaction is the question is:
$C{{H}_{4}}+{{H}_{2}}O\to CO+{{H}_{2}}$
On balancing this equation, we get:
$C{{H}_{4}}+{{H}_{2}}O\to CO+3{{H}_{2}}$
We can clearly see that one mole of methane will produce 3 moles of hydrogen gas. To find the volume of hydrogen gas produced, we can use the formula:
$\dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{{{V}_{1}}}{{{V}_{2}}}$
Where n and V are the numbers of moles are volume of the compounds respectively. Let 1 represent methane and 2 represents hydrogen gas. The volume of methane is 100 $c{{m}^{3}}$. Putting the values, we get:
$\dfrac{1}{3}=\dfrac{100}{{{V}_{2}}}$
${{V}_{2}}=300\text{ c}{{\text{m}}^{3}}$
Therefore, 300 $c{{m}^{3}}$ of hydrogen gas will be produced.

Hence the correct answer is an option (c).

Note: If the reaction is not given in the question, and only mass is given then you can calculate the number of moles by dividing the given mass by the molecular mass of the compound and then place it in the law.