1 g of Mg is burnt in a closed vessel containing 0.5 g of `O_2`. Which reactant is limiting reagent and how much of the excess reactant will be left?

Explanation

Carbon burns in oxygen according to the following equation.
`underset(12.01 g)C + underset(32.0 g)(O_(2)) to underset(12.01 + 32.0 = 44.01 g)(CO_(2))`
`therefore` 12.01 g of carbon require for combustion, = 32.0 g
`therefore 1.0` g of carbon will require `=(32.0)/(12.01) xx 1.0 = 2.7 g`
Hence, the oxygen present in the closed vessel is much less than the required amount. The amount of `O_(2)` will thus be completely consumed and carbon will be left in excess. Thus, `O_2` is the limiting reagent in this case.
`therefore 32.0` g of oxygen burn carbon = 12.01 g
1.5 g of oxygen will burn carbon `=12.01/32.0 xx 1.5 = 0.56 g`
Hence, carbon left unreacted = `1.0 - 0.56 = 0.44` g Now, `therefore 32.0` g of oxygen combine with carbon to give `CO_2 = 44.01` g
`therefore` 1.5 g of oxygen will give `CO_(2) = (44.01)/(32.0) xx 1.5 = 2.1 g`
Hence, carbon is left in excess in the given reaction. The amount of excess reactant is 0.44 g and the mass of the product `CO_2` formed in the reaction is 2.1 g