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Question: The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to form di...

The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to form dihydrogen. What volume of dihydrogen at 20oC{20^{\text{o}}}{\text{C}} and one bar will be released when 0.150.15g of aluminium reacts?

Explanation

Solution

The question is based on the stoichiometric reaction of aluminium with caustic soda or sodium hydroxide. We shall calculate the moles of hydrogen formed from the reaction equation and substitute the values in the ideal gas equation.
Formula Used:
moles = massmolar mass{\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}
The combined gas law equation, PV = nRT{\text{PV = nRT}}
Where, P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant in litre atm, and T is the temperature of the gas in Kelvin Scale.

Complete step by step answer:
The reaction can be represented as:
2Al + 2NaOH + 2H2O→2NaAlO2 + 3H2{\text{2Al + 2NaOH + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2NaAl}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}
2 moles of aluminium reacts with 2 moles of sodium hydroxide to form 3 moles of dihydrogen gas as per the stoichiometry of the reaction.
So, 1 mole aluminium gives 32\dfrac{3}{2} moles of hydrogen
Atomic weight of aluminium = 26  g/mol26\;{\text{g/mol}}
Number of moles of aluminium = weightmol. weight\dfrac{{{\text{weight}}}}{{{\text{mol}}{\text{. weight}}}}= 0.1526=5.56×10−3\dfrac{{0.15}}{{26}} = 5.56 \times {10^{ - 3}} moles.
No. of moles of hydrogen =32×5.56×10−3\dfrac{3}{2} \times 5.56 \times {10^{ - 3}}= 8.33×10−38.33 \times {10^{ - 3}} moles.
Now, to find the volume of the gas, we use the combined gas equation, considering dihydrogen to behave ideally.
PV = nRT{\text{PV = nRT}}
⇒V = nRTP\Rightarrow {\text{V = }}\dfrac{{{\text{nRT}}}}{{\text{P}}}
Substituting the values:
⇒V = 8.33×10−3×0.0821×2931\Rightarrow {\text{V = }}\dfrac{{8.33 \times {{10}^{ - 3}} \times 0.0821 \times 293}}{1} = 0.2030.203 litres
Or, V = 0.2030.203 litres.

So, the amount of chlorine liberated from 0.150.15 g of aluminium is 0.2030.203 litres.

Note:
The Combined gas law equation is valid only for ideal gases in which the pressure exerted by the molecules of the gas is considered to be very low and the volume of the gas molecules is considered to be negligible in comparison to the volume of the gas. When not mentioned in the question, we always assume the gas as an ideal gas.