Question

QUESTION:

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

(CLASS 11th, NCERT, CHEM.PART1, Pg.no.158, EXERSISE 5.6)

•No need of copied answer.
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Answer 1

Al react with caustic soda to produce dihydrogen. reaction will be

2Al + 2NaOH + 2H2O -->2NaAlO2 + 3H2

here we see that, 2 mol of Al produce 3 mole of H2

∴ 2 × 27 g of Al produce 3 × 2 g of H2 .

∴ 54 g of Al produce 6g of H2.

∴ 1 g of Al produce 6/54 = 1/9 g of H2.

∴ 0.15g of Al produce 0.15/9 g of H2

so, number of mole of H2 produce = weight of HE produce/molar weight of H2

= (0.15/9)/2

= 0.15/18 = 0.00833 mol

now, use formula, PV = nRT

Here, P = 1 bar or 0.987 atm

T = 20°C or 293K

n = 0.00833 mol

R = 0.0821 atm dm³/K/mol

V = ?

V = 0.00833 × 0.0821 × 293/0.987

= 0.203 dm³

we know, 1 dm³ = 1000 mL

so, Volume of H2 = 1000× 0.203mL = 203mL

Answer 2

$\blue{ \underline{ \boxed{ \sf { \bigstar \: Question}}}}$

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

$\red { \underline{ \boxed{ \sf { \bigstar \: Answer}}}}$

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

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

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

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."