Carbon burns in oxygen as shown by the chemical equation: `2C(s) + O_(2)(g) to 2CO(g)`.
A reaction is carried out starting with 12 g carbon and 48 g oxygen.
Which reactant will be in excess at the end of the reaction?

Answer

1

Explanation

Given mass of carbon = 12 g and mole of carbon = 12 g/12 g = 1 mol
Moles of oxygen ` = 48 g//32 g " mol"^(-1) = 1.5` mol
`{:(2C(s),+,O_(2)(g),to,2CO(g)),("2 mol ",,"1 mol ",,"2 mol"),("1 mol",,1/2"mol",,"1 mol"):}`
That is , 1 mol carbon will combine with `0.5 ` mol oxygen to produce 1 mole of carbon monoxide. Therefore,
Oxygen will be in excess according to the given reaction. `1.5` mol `O_(2) - 0.5 " mol " O_(2) = 1.0 ` mol excess of `O_(2)` . Thus, carbon is the limiting reactant.