Question

Find the volume of hydrogen which would be evolved by STP by reacting zinc with $ 50{\text{ mL}} $ of $ 40\% $ pure dilute $ {H_2}S{O_4} $ (density $ = 1.3\;gm{L^{ - 1}} $ ). Report your answer in litres and round off to the nearest integer.

Answer 1

Hint : In a chemical reaction, the reactant which is completely consumed is known as limiting reagent while the reactant which will be left after the reaction is complete is known as excess reagent. The limiting reagent plays a major role in deciding when the chemical reaction will stop and the amount of product formed.

Complete Step By Step Answer:
Reaction of zinc with dilute sulphuric acid takes place as follows:
 $ Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2} $
As the mass of zinc is not given in the question so we will assume that the limiting reagent for the given reaction will be $ {H_2}S{O_4} $ . As per question, the given data is as follows:
Volume of $ {H_2}S{O_4} = 50{\text{ mL}} $
Density of $ {H_2}S{O_4} = 1.3\;gm{L^{ - 1}} $
We know that the density of a molecule is the ratio of its mass to its volume. So, mass of $ {H_2}S{O_4} $ will be as follows:
Density $ = \dfrac{{{\text{mass}}}}{{{\text{volume}}}} $
Substituting values:
 $ \Rightarrow $ Mass $ = 50 \times 1.3 $
 $ \Rightarrow $ mass $ = 65g $
But in the reaction, only $ 40\% $ of dilute $ {H_2}S{O_4} $ is consumed. So, the actual mass of sulphuric acid consumed in the reaction will be as follows:
m $ = {\text{Mass}} \times \dfrac{{40}}{{100}} $
 $ \Rightarrow m = 65 \times 0.4 $
 $ \Rightarrow m = 26g $
Therefore, number of moles of sulphuric acid consumed in the reaction will be as follows:
n $ = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}} $
Molar mass of $ {H_2}S{O_4} = 98\;gmo{l^{ - 1}} $
Substituting values, then the number of moles of $ {H_2}S{O_4} $ will be as follows:
 $ \Rightarrow n = \dfrac{{26}}{{98}} $
 $ \Rightarrow n = 0.265\;{\text{moles}} $
Now as per given reaction,
 $ 1 $ mole of $ {H_2}S{O_4} $ reacts with zinc to form $ \Rightarrow 1 $ mole of $ {H_2} $
Therefore, $ 0.265 $ moles of $ {H_2}S{O_4} $ will react with zinc to form $ \Rightarrow 1 \times 0.265 = 0.265 $ moles of $ {H_2} $
We know that, at STP conditions, $ 1 $ mole of a gas occupies $ 22.4{\text{ L}} $ of volume. So, volume occupied by $ 0.265 $ moles of $ {H_2} $ will be as follows:
Volume $ = 0.265 \times 22.4 $
 $ \Rightarrow V = 5.93{\text{ L}} $
 $ \Rightarrow V \approx 6\;{\text{L}} $
Hence, the volume of hydrogen gas produced under given reaction conditions $ = 6{\text{ L}} $ .

Note :
It is important to note that when zinc reacts with dilute sulphuric acid at room temperature then formation of zinc sulphate takes place along with the removal of hydrogen gas. But when the zinc metal is heated with concentrated sulphuric acid, then formation of zinc sulphate takes place along with the removal of sulphur dioxide and water.