1.0 g of Mg is burnt with 0.28 g of `O_2` in a closed vessel . Which reactant is left in excess and how much ?

Options

A. Mg, 5.8 g
B. Mg, 0.58 g
C. `O_(2)`, 0.24 g
D. `O_(2)`, 2.4 g

Answer

B

Explanation

`2Mg + O_(2) to 2MgO`
Moles of Mg `=1.0/24 = 0.042`
Moles of `O_(2) =0.28/32 = 0.00875`
1 mol of `O_(2)` requires Mg =2 mol
0.00875 mol of `O_(2)` requires Mg `=2 xx 0.00875= 0.0175`
`therefore` Mg is in excess.
Moles of Mg left in excess `=0.042 - 0.0175 = 0.0245`
Mass of Mg left in excess =0.0245 x 24 = 0.58 g