The volume of `H_2` evolved at STP when `0.9 g` of Al(molar mass : `27 g mol^(-1)`) is dissolved in excess of dilute `H_(2)SO_(4)` is

Options

A. `0.58 L`
B. `1.12 L`
C. `2.40 L`
D. `2.9 L`

Answer

B

Explanation

The reaction is `2Al + 3 H _(2) SO _(4) to Al_(2) (SO _(4)) _(3) + 3 H _(2)`
Amount of `Al = (( 0. 9g)/( 27 g mol ^(-1))) = (1 )/(30) mol,` Amount of `H_(2)` evolved `=((3)/(2)) ((1)/(30) mol) = (1)/(20) mol `
Volume of `H_(2)` evolved `= ((1)/(20) mol) (22.4 L mol ^(-1)) = 1. 12 L`