Question

The weight of zinc required for the liberation of 10 g of hydrogen gas on reaction with $\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ is:

Answer 1

Stoichiometry states the relationship between the numbers of moles of reactant to the number of moles of the product. Here, from stoichiometric relation, the one mole of zinc metal reacts with the one mole of the sulphuric acid to produce one mole of zinc sulfate and hydrogen gas. Since a mole is a ratio of weight to the molar mass, we can say that,
$1\text{ mole of Zn = 1 mole of }{{\text{H}}_{\text{2}}}\text{ gas}$

Complete step by step answer:
Stoichiometry establishes the exact quantitative relationship between the number of moles of reactants and products in the chemical reaction. To exhibit a chemical reaction, the reaction must be balanced in other words each atom on the reactant and product side must be the same. The stoichiometry works on the principle of conservation of mass.
Let us consider a reaction of metal with acid.
The metal reacts with the acid to form salt and liberates hydrogen gas. Not all metals react in the same way to acids. the general reaction between then metal (M) and acid can be depicted as follows:

$\text{ Metal + Acid }\to \text{ Salt + Hydrogen gas (}\uparrow \text{)}$
The zinc$\text{ Zn }$ metal reacts with the sulphuric acid$\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }$, the reaction proceeds with the formation of zinc sulfate $\text{ ZnS}{{\text{O}}_{\text{4}}}\text{ }$and liberates the hydrogen gas $\text{ }{{\text{H}}_{\text{2}}}$. The reaction can be written as:

$\text{ Zn + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\to \text{ ZnS}{{\text{O}}_{\text{4}}}\text{ + }{{\text{H}}_{\text{2}}}$
From stoichiometry, one mole of zinc produces one mole of hydrogen gas.

Therefore, the above relation can be written as:
$1\text{ mole of Zn = 1 mole of }{{\text{H}}_{\text{2}}}\text{ gas}$
As the atomic weight of $\text{Zn}$ is $\text{65}\text{.38 g }$ and atomic weight ${{\text{H}}_{\text{2}}}$ is$\text{2 g }$.
Then,
$\text{10 g of hydrogen = }\dfrac{10}{2}\text{ = 5 mole}$
We know that one mole of hydrogen gas is produced from the one mole of zinc.
Then, 5 moles of hydrogen gas are produced from the x number of moles.
$10\text{ g}$Of hydrogen gas contain,
$\text{65}\text{.38 g of Zn = 2 g of }{{\text{H}}_{\text{2}}}\text{ gas}$

Now, for the liberation of$\text{ 10 g }$ ${{\text{H}}_{\text{2}}}$, the amount of zinc required can be written as:

& \begin{matrix} \text{1 mole of Zn} & \text{=} & \text{1 mole of Zn} \\\ \text{x mole of Zn} & \text{=} & \text{5 moles of }{{\text{H}}_{\text{2}}}\text{ gas} \\\ \text{Therefore,} & {} & {} \\\ \end{matrix} \\\ & \begin{matrix} \text{x mole of Zn } & \text{=} & \dfrac{\text{5 }\\!\\!\times\\!\\!\text{ 1}}{\text{1}} \\\ \end{matrix}\text{mole = 5 mole of zinc} \\\ \end{aligned}$$ We know that, $\text{ mole = }\dfrac{\text{mass}}{\text{molar mass}}$ Therefore, calculate the weight of the zinc required. $\begin{aligned} & \text{ 5 mole of Zn = }\dfrac{\text{m}}{\text{65}\text{.38 g mo}{{\text{l}}^{\text{-1}}}} \\\ & \Rightarrow \text{m = }\left( \text{5 mole of Zn} \right)\text{ }\\!\\!\times\\!\\!\text{ }\left( \text{65}\text{.38 g mo}{{\text{l}}^{\text{-1}}} \right) \\\ & \therefore \text{ m = 326}\text{.9 g} \\\ \end{aligned}$ Therefore, the total amount of zinc required for the liberation of $$10\text{ g}$$ hydrogen gas is $$\text{326}\text{.9 g}$$. **Note:** Three types of relationships can be established between the chemical substances involved in a chemical reaction. These are 1) Weight-weight relationship of reactant and product 2) weight-volume relationship where one reactant is at least in a gas state 3) Volume – volume relationship in case all reaction takes place in a gaseous state. The above reaction is an example of the weight-weight relationship between reactant and product.