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

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We have the chemical reaction:

$2 Al +2NaOH +H_2O \rightarrow 2NaAlO_2+3H_{2}$

(Where dihydrogen is being produced.)

So, at $20^{\circ}C$ and $1\ bar$ pressure, the volume of gas that will be released when $0.15g$ of aluminium reacts.

As we can see from the reaction equation that 2 moles of aluminium produce 3 moles of dihydrogen.

i.e., $2\times 27g$ reacts to give $3\times 22.4Litres$

At STP condition

$273.15K$ and $1atm$ ,

$54g(2\times 27g)$ of Al gives $3\times 22400mL$ of $H_{2}$ .

Therefore $0.15g$ of Al will give $= \frac{3\times22400\times0.15}{54}mL$ of $H_{2}$

i.e., $186.67mL$ of $H_{2}$ .

At STP condition:

$273.15K$ and $1atm$ ,

Now, $P_{1} =1atm$ , $V_{1} = 186.67mL$ , $T_{1} = 273.15K$

Then we assume the volume of dihydrogen be $V_{2}$ at the pressure $P_{2} = 0.987atm$

(Since 1 bar = 0.987atm) and temperature $T_{2} = 20^{\circ}C$

$\implies (273.15+20)K = 293.15K$

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

$V_{2}= \frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}$

$= \frac{1\times186.67\times293.15}{0.987\times273.15}$

$= 202.98mL\approx 203mL$

Therefore, 203mL of dihydrogen will be released.

Posted by

Divya Prakash Singh

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