JEE Advance - Chemistry (2020 - Paper 1 Offline - No. 16)
Aluminium reacts with sulphuric acid to form aluminium sulphate and hydrogen. What is the volume of hydrogen gas in litre (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 g mol$$-$$1, R = 0.082 atm L mol$$-$$1 K$$-$$1)
(Use molar mass of aluminium as 27.0 g mol$$-$$1, R = 0.082 atm L mol$$-$$1 K$$-$$1)
Answer
6.15
Explanation
Aluminium reacts with sulphuric acid to form aluminium sulphate and hydrogen.
$$\mathop {2Al}\limits_{\left( {{{5.4} \over {27}} = 0.2\,mol} \right)} + \mathop {3{H_2}S{O_4}}\limits_{\left( {{{50 \times 5} \over {1000}} = 0.25\,mol} \right)} \buildrel {} \over \longrightarrow A{l_2}{(S{O_4})_3} + 3{H_2}$$
H2SO4 is limiting reagent and moles of H2(g) produced = 0.25 mol
Using ideal gas equation,
pV = nRT
$$ \Rightarrow $$ $$V = {{0.25 \times 0.082 \times 300} \over {1\,atm}} = 6.15\,L$$
$$\mathop {2Al}\limits_{\left( {{{5.4} \over {27}} = 0.2\,mol} \right)} + \mathop {3{H_2}S{O_4}}\limits_{\left( {{{50 \times 5} \over {1000}} = 0.25\,mol} \right)} \buildrel {} \over \longrightarrow A{l_2}{(S{O_4})_3} + 3{H_2}$$
H2SO4 is limiting reagent and moles of H2(g) produced = 0.25 mol
Using ideal gas equation,
pV = nRT
$$ \Rightarrow $$ $$V = {{0.25 \times 0.082 \times 300} \over {1\,atm}} = 6.15\,L$$
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