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The drain cleaners, orainer contain small bits of \mathrm{Al} which reacts with caustic soda to produce dihy drogen. what volume of dihydrogen at \mathrm{20^{\circ}C} and 1 bar will be released when 0.16 g Al reacts ______

Option: 1

250 ml


Option: 2

203 ml


Option: 3

400 ml


Option: 4

403 ml


Answers (1)

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The following reaction -

\mathrm{\begin{aligned} & 2 \mathrm{Al}(s)+2 \mathrm{NaOH}(\mathrm{a} q)+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+3 \mathrm{H}_2 \\ & 2 \times 2+g \\ & \end{aligned}}

According ti this reaction , \mathrm{54 \mathrm{~g} \text { of } \mathrm{Al}=3 \mathrm{~mol}} of \mathrm{H_{2}} 

\mathrm{\therefore 0.15 \mathrm{~g} \text { of } \mathrm{Al}=\frac{3}{54} \times 0.15=8.33 \times 10^{-3} \mathrm{~mol} \text { of } \mathrm{H}_2}

To calculate volume of the liberated \mathrm{\mathrm{H}_2} gas, we apply the ideal gas equation , \mathrm{PV=nRT}

Given : \mathrm{P=1 \mathrm{bar} \text { and } T=(273+20) k=2.93 \mathrm{k}}

The number of moles of liberated \mathrm{H_{2}} gas \mathrm{\left.(n)=2.33 \times 10^{-9} \mathrm{mol}\right)}

 \mathrm{\begin{aligned} & \therefore V=\frac{n R T}{P}=\frac{8.33 \times 10^{-3} \times 0.0821 \times 293}{0.987}=0.203 \mathrm{~L} \\ & =203 \mathrm{ml} \\ & \end{aligned}}

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Ritika Jonwal

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