The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at and one bar will be released when \[0.15{\text{ }}g\]of aluminium reacts?
\[
  {A.{\text{ }}204{\text{ }}mL} \\
  {B.{\text{ }}200{\text{ }}mL} \\
  {C.{\text{ }}203\;mL} \\
  {D.{\text{ }}400{\text{ }}mL}
\]

Hint: First calculate the parameters at STP and then use the ideal gas equation to find it at given requirements.
Formula Used:
 Ideal Gas Equation: $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$

Complete step by step solution:
Let’s start with writing the chemical reaction equation between the aluminium and caustic soda.
\[2Al{\text{ }} + {\text{ }}2NaOH{\text{ }} + {\text{ }}2{H_2}O{\text{ }} \to {\text{ }}2NaAl{O_2} + {\text{ }}3{H_2}\]
From this equation we get that \[2{\text{ }}moles\] of Aluminium is producing \[3{\text{ }}moles\] of dihydrogen gas.
At STP (Standard temperature and pressure) \[1{\text{ }}mole\]of dihydrogen gas is having the volume of\[22.4{\text{ }}litres\].
So, \[2{\text{ }}X{\text{ }}27\](molar mass of aluminium) gives us \[3{\text{ }}X{\text{ }}22.4\]litres of dihydrogen.
Which means \[54{\text{ }}grams\]of aluminium gives \[3{\text{ }}X{\text{ }}22.4{\text{ }}L\]of dihydrogen at STP.
Now, we are having \[0.15g\] of aluminium, so, how much litres of \[{H_2}\]will it produce at STP.
$\dfrac{{3 \times 22.4 \times 0.15}}{{54}}$= \[0.186{\text{ }}L{\text{ }}of{\text{ }}{H_2}at{\text{ }}STP.\]
Now for calculating at and \[1{\text{ }}bar\]pressure we apply the ideal gas equation.
Ideal Gas Equation: $\dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{T}}_1}}} = \dfrac{{{{\text{P}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}}$
We are having\[{P_1} = {\text{ }}1atm,{\text{ }}{V_1} = 0.187L,{\text{ }}{T_1} = {\text{ }}273K{\text{ }}and{\text{ }}{P_2} = {\text{ }}1bar = 0.987{\text{ }}atm,{\text{ }}{T_2} = {\text{ }}293K\], \[{V_2}\]we have to find
So, putting these values in the equation we get
$\dfrac{{1 \times 0.187}}{{273}} = \dfrac{{0.987 \times {V_2}}}{{293}}$
 ${V_2} = \dfrac{{293}}{{0.987}} \times \dfrac{{0.187}}{{273}}$
\[{V_2} = {\text{ }}0.2030{\text{ }}L\]which is equal to \[203mL\]

So, the answer to this question is C. \[203mL\]

Note: We must know that ideal Gas Equation has a wide range of applications, ranging from solving problems like shown above to solving some complicated equations. But, nothing is ideal in nature. We can only bring things close to ideal but not ideal. Because of this there are some constants that are used to make this simple looking equation a little bit complex but suitable for non-ideal conditions.