Which one of the following gives highest volume of `O _(2)` at STP on complete decompositon ?

Options

A. 2 mL of 100 vol. `H _(2) O_(2)`
B. 500 mL of 30 vol. `H _(2) O _(2)`
C. 1 L of 10 vo. `H_(2)O_(2)`
D. 100 mL of 20 vol.`H_(2)O_(2)`

Answer

B

Explanation

`H_(2)O_(2)` on decomposition gives `H _(2)O and O _(2)` i.e., `2H _(2)O_(2) hArr 2H_(2) O+O _(2)`
`1. 2mL of 100` vol `H _(2) O _(2)`
At STP, 100 vol `H _(2) O _(2) implies`
1 mL of `H _(2)O_(2)-= 100 mL of O _(2)`
Then, `2 mL of H _(2)O _(2) -= ( 2xx 100)/(1) = 200 mL of O _(2)`
2. 500 mL of 30 vol `H _(2)O _(2)`
At STP, 30 vol `H _(2) O _(2) implies`
`1 mL of H _(2)O _(2) -= 30 mL of O _(2)`
Then, ` 500 mL of H _(2) O _(2) -= (500 xx 30)/(1) = 15000 mL of O _(2)`
`3. 1 L of 10 vol. `H _(2) O _(2)`
At STP, 10 vol `H _(2) O _(2) implies`
`1mL of H _(2) O _(2) -= 10 mL of O _(2)`
Then, 1000 mL or 1 L of `H _(2) O _(2) -= (100 0 xx10)/( 1) = 10000mL of O _(2)`
`4. 100 mL of 20 vol `H _(2) O _(2)`
At STP, 20 vol `H _(2)O _(2) implies`
1 mL of `H _(2)O _(2) -= 20 mL of O _(2)`
Then, `100 mL of H _(2)O_(2) -= ( 100 xx 20)/(1) = 2000 mL of O _(2)`
`therefore 500 mL of 30 ` vol `H _(2) O _(2)` gives highest volume of `O _(2)` at STP.