Calculate the volume of air containing 21% by volume of oxygen at NTP required to convert 294mL of `SO_(2)` into `SO_(3)` under the same conditions.

Answer

`700cm^(3)`

Explanation

`underset(2xx27g)(2Al)+2NaOH+2H_(2)Orarr2NaAlO_(2)+underset(3xx22400"mL at STP")(3H_(2))`
`therefore" "H_(2)" produced at STP from 0.15 g Al"=(3xx22400)/(54)xx0.15" mL = 186.7 mL"`
`underset("(STP conditions)")((P_(1)V_(1))/(T_(1)))=underset("(Required conditions)")((P_(2)V_(2))/(T_(2)))`
`"i.e., "(1atmxx186.7mL)/(273K)=(0.987atm xxV_(2))/(293K)" (1 bar = 0.987 atm)"`
`"or "V_(2)=203.0mL`