The volume of oxygen evolved at STP by decomposition of `0.68` g '20 volume ' hydrogen peroxide solution is

Options

A. 2.24 mL
B. 22.4 mL
C. 224 mL
D. 2240 mL

Answer

C

Explanation

` underset( = 2 xx 34 = 68 g) underset(2 "mol")(2H_(2)O_(2)) to 2H_(2) O + underset(22400 mL)underset(1 "mol")(O_(2))`
`because` At STP , 68 g of `H_(2)O_(2)` produced
`O_(2) = 22400` mL
`therefore` At STP , `0.68 g H_(2)O_(2)` will produce
`O_(2) = (22400 xx 0.68)/(68) = 224` mL