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?

Explanation

Step I. Calculation of the volume of hydrogen released under N.T.P. conditions
The chemical equation for the reaction is :
`underset(2xx27=54g)(2Al+2NaOH)+2H_(2)O+2H_(2)Orarrunderset("Sod. meta aluminate")(2NaAlO_(2))+underset(3xx22400 mL)(3H_(2))`
`54g` of Al at N.T.P release `H_(2)` gas ` = 3 xx 22400 mL`
`0.15 g` of Al at N.T.P. release `H_(2)` gas `((3 xx 22400 mL) xx (0.15 g))/((54 g)) = 186.7 mL`.
Step II Calculating of volume of hydrogen released at `20^(@)C` and 1 bar pressure.
`{:("N.T.P. Conditions",,"Given Conditions"),(V_(1)=186.67mL,,V_(2)=?),(P_(1)=1.013"bar",,P_(2)=1"bar"),(T_(1)=0+273=273K,,T_(2)=20+273=293K):}`
According to gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(P_(2)T_(1))`
By substituting the values, `V_(2) = ((1.013 "bar" ) xx (186.7 mL) xx (293 K))/((1 "bar") xx (273 K)) = 203 mL`.