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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} = 186.67mLT_{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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