1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24, O = 16)
(a) Mg, 0.16 g
(c) Mg, 0.44 g
(b) O2, 0.16 g
(d) O2, 0.28 g
Answer - (a) Mg, 0.16g
Solution -
2Mg + O2 → 2MgO
no of moles of Mg = 1/24 = 0.0416 moles
no of moles of O2 = 0.56/32 = 0.0175 moles
0.0416/2 = 0.0208 > 0.0175
thus O2 is limiting reagent
Reactants | Mg | O2 |
---|---|---|
Initial moles | 0.0416 | 0.0175 |
Final moles | 0.0416 - (2 x0.0175) = 0.0066 | 0.0175 - (1 x 0.0175) = 1 |
mass of Mg = 0.0066 x 24 = 0.16g