Question

The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and 1 bar will be released when 0.15 g of aluminium reacts?

Answer 1

"Aluminium" reacts with "caustic soda" in accordance with the "reaction".

The reaction of aluminium with caustic soda can be represented as follows

$\begin{matrix} 2Al \\ \left( 2\quad \times \quad 27 \right) g \end{matrix}\quad +\quad 2NaOH\quad +\quad 2{ H }_{ 2 }O\quad \longrightarrow \begin{matrix} 2NaAl{ O }_{ 2 } \\ \left( 3\quad \times \quad 22400 \right) ml \end{matrix}\quad +\quad 3{ H }_{ 2 }$

Therefore, volume of hydrogen at STP released

When 0.15 g of Al reacts

$V\quad =\quad \frac { 0.15\quad \times \quad 3\quad \times \quad 22.4 }{ 54 } \quad =\quad 187\quad ml$

From given

${ P }_{ 1 }$ = 1 bar

${ P }_{ 2 }$ = 1 bar

${ T }_{ 1 }$ = 273 K

${ T }_{ 2 }$= 20 + 273 = 293 K

$V_{ 1 } =$ 187 ml  

${ V }_{ 2 }$= ?

When pressure is held constant, then $V_{ 2 }\quad =\quad \frac { { P }_{ 1 }{ V }_{ 1 }{ T }_{ 2 } }{ { P }_{ 2 }{ T }_{ 1 } }$

Or

${ V }_{ 2 }\quad =\quad \frac { 1\quad \times \quad 187\quad \times \quad 293 }{ 0.987\quad \times \quad 273 } \quad =\quad 201\quad ml$

Therefore, 201 ml of dihydrogen will be formed."

Answer 2

"Aluminium" reacts with "caustic soda" in accordance with the "reaction".

The reaction of aluminium with caustic soda can be represented as follows

\begin{lgathered}\begin{matrix} 2Al \\ \left( 2\quad \times \quad 27 \right) g \end{matrix}\quad +\quad 2NaOH\quad +\quad 2{ H }_{ 2 }O\quad \longrightarrow \begin{matrix} 2NaAl{ O }_{ 2 } \\ \left( 3\quad \times \quad 22400 \right) ml \end{matrix}\quad +\quad 3{ H }_{ 2 }\end{lgathered}

2Al

(2×27)g

+2NaOH+2H

2

O⟶

2NaAlO

2

(3×22400)ml

+3H

2

Therefore, volume of hydrogen at STP released

When 0.15 g of Al reacts

V\quad =\quad \frac { 0.15\quad \times \quad 3\quad \times \quad 22.4 }{ 54 } \quad =\quad 187\quad mlV=

54

0.15×3×22.4

=187ml

From given

{ P }_{ 1 }P

1

= 1 bar

{ P }_{ 2 }P

2

= 1 bar

{ T }_{ 1 }T

1

= 273 K

{ T }_{ 2 }T

2

= 20 + 273 = 293 K

V_{ 1 } =V

1

= 187 ml

{ V }_{ 2 }V

2

= ?

When pressure is held constant, then V_{ 2 }\quad =\quad \frac { { P }_{ 1 }{ V }_{ 1 }{ T }_{ 2 } }{ { P }_{ 2 }{ T }_{ 1 } }V

2

=

P

2

T

1

P

1

V

1

T

2

Or

{ V }_{ 2 }\quad =\quad \frac { 1\quad \times \quad 187\quad \times \quad 293 }{ 0.987\quad \times \quad 273 } \quad =\quad 201\quad mlV

2

=

0.987×273

1×187×293

=201ml

Therefore, 201 ml of dihydrogen will be formed."