The drain cleaner Drainex contains small bits of aluminium which react with caustic soda to produce hydrogen What volume of hydrogen at `20^(@)C` aand one bar will be released when `0.15 g` of aluminium reacts ? .

Explanation

`2 Al + 2NaOH + 2H_2O to 2NaAlO_2 + 3H_2`
`2 xx 26.98 " " 3 xx 22.4`
= `53.96 g " " =67.2L at S.T.P`.
From the above, it is clear that 53.96 g of Al react with NaOH to form 67.2 L of `H_2` at S.T.P.
`:. " Volume of " H_2` at S.T.P. formed by 0.15 g of Al
`=(67.2)/(53.96)xx0.15=0.1868 L`
Suppose, the volume of this hydrogen at `20^@C (293 K)` and 1 bar (0.987 atm) is `V_2`. According to the gas equation,
`(P_1 V_1)/T_1 = (P_2 V_2)/T_2`
or `(1xx0.1868)/273 = (0.987xxV_2)/293`
or `V_2 = (0.1868xx293)/(0.987xx273) = 0.2031 L = 203.1 mL`
`293 or V2 = 0.1868 x 293 - 2031 = 203.1 mL`