Question

Aluminium reacts with sulphuric acid to form aluminium sulphate and hydrogen. What is the volume of hydrogen gas in litres (L) produced at 300 K and 1.0 atm pressure, when 5.4 g of aluminium and 50.0 mL of 5.0 M sulphuric acid are combined for the reaction? (Use molar mass of aluminium as 27.0 $ gmo{l^{ - 1}} $ , R= 0.082 $ atm\;.L.mo{l^{ - 1}}{K^{ - 1}} $ )

Answer 1

Hint :We know that when 2 moles of Aluminium will react with 3 moles of sulphuric acid and give us one mole of aluminium sulphate and 3 moles of hydrogen gas.
This can be represented in the form of the given equation:
 $ \Rightarrow 2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2} $

Complete Step By Step Answer:
One mole of aluminium contains 27 grams of aluminium. We are given 5.4 grams of aluminium which means that we have 0.211 moles of aluminium.
This is found out by the following equation:
 $ \Rightarrow No.{\text{ }}of{\text{ }}moles = \dfrac{{Mass{\text{ of the given compound}}}}{{Molar{\text{ mass of the compound}}}} $
 $ \Rightarrow No.{\text{ }}of{\text{ }}moles = \dfrac{{5.4g}}{{27gmo{l^{ - 1}}}} $
 $ \Rightarrow No.{\text{ }}of{\text{ }}moles = 0.2{\text{ }}moles $
Now we have to find out the moles of sulphuric acid that we have
 $ \Rightarrow No.{\text{ }}of{\text{ }}moles{\text{ of }}{{\text{H}}_2}S{O_4} = \dfrac{{5 \times 50}}{{1000}} = 0.25{\text{ moles}} $
Thus we can say that we have 0.2 moles of aluminium and 0.25 moles of sulphuric acid.
Since 2 moles of Aluminium react with 3 moles of sulphuric acid, we can say that
0.2 moles of aluminium react with 0.3 moles of sulphuric acid
Thus we can conclude that sulphuric acid will act as the limiting reagent.
Total moles of hydrogen gas formed will be $ \dfrac{{3 \times 0.25}}{3} $ = 0.25 moles of hydrogen gas
We also know by ideal gas equation:
 $ \Rightarrow {\text{PV = nRT}} $
Thus we can say that
 $ \Rightarrow {\text{V = }}\dfrac{{{\text{nRT}}}}{P} $
 $ \Rightarrow {\text{V = }}\dfrac{{0.25 \times 0.082 \times 300}}{1} $
 $ \Rightarrow {\text{V = 6}}{\text{.15 Litres}} $
Thus the amount of hydrogen gas produced in this reaction is 6.15 litres.

Note :
In questions like these, always make sure to calculate the moles of the given reactant and then proceed in finding the quantity asked in the question. According to the law of constant proportions, the chemical compounds are made up of elements that are present in a fixed ratio by mass. This law is being manipulated in these problems