1.0 g of magnesium is burnt with 0.56 g `O_(2)` in a closed vessel. Which reactant is left in excess and how much?

Options

A. Mg, 0.16 g
B. `O_(2)`, 0.16 g
C. Mg, 0.44 g
D. `O_(2)`, 0.28 g

Answer

A

Explanation

Moles of Mg `=1/24 = 0.0417` mol
Moles of `O_(2) =0.56/32 = 0.0175` mol
`underset(1 mol)(Mg) + underset(0.5 mol)(1/2 O_(2)) to underset(1"mol")(MgO)`
`O_(2)` is limiting reagent.
0.5 mol of `O_(2)` reacts with Mg to form 1 mol of MgO
0.0175 mol of `O_(2)` react iwth Mg `=(1 xx 0.0175)/0.5`
Moles of Mg left unreacted `=0.0417 - 0.035`
`=6.7 xx 10^(-3)` mol
Amount of Mg left unreacted `=6.7 xx 10^(-3) xx 24`
=0.1608 g