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

`underset(=54)underset(2xx27)(2Al)+2NaOH+2H_(2)Orarrunderset(3xx"22400 mL")(2NaAlO_(2)+2H_(2))`
It is known that 1 mol of any gas at S.T.P. i.e., `0^(@)C` and 1 atm (or 1.01325 bar) pressure occupies 22400 mL
`"0.15 g of Al will give "H_(2)=(3xx22400)/(54)xx0.15`
`=186.7mL`
So, 186.7 mL of `H_(2)` will be released at 1.01325 bar pressure and 273 K. To calculate volume of `H_(2)` at `20^(@)C` and 1 bar pressure,
`p_(1)="1.01325 bar "p_(2)="1 bar"`
`V_(1)="186.7 mL "V_(2)=?`
`T_(1)=273K" "T_(2)=273+20=293K`
`(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))or V_(2)=(p_(1)V_(1)xxT_(2))/(p_(2)xxT_(1))`
`therefore" "V_(2)=("1.01325 bar"xx"186.7 mL"xx293K)/("1 bar "xx"273 K")`
`=203.03mL`