12 g of Mg (at. Mass 24) will react completely with acid to give
A. One mole of `H_(2)`
B. `1//2` mole of `H_(2)`
C. `2//3` mole of `O_(2)`
D. Both `1//2` mol of `H_(2)` and `1//2` mol of `O_(2)`
A. One mole of `H_(2)`
B. `1//2` mole of `H_(2)`
C. `2//3` mole of `O_(2)`
D. Both `1//2` mol of `H_(2)` and `1//2` mol of `O_(2)`
Correct Answer – B
`Mg^(+2) equiv H_(2)`
`n=(12 gm)/(24 gm)=1/2` mole of `H_(2)`