Question

A `5*g` mass of magnesium metal reacts with a `4.0*g` mass of oxygen gas. Which reagent is in stoichiometric excess?

Answer 1

We follow the stoichiometric equation:

`2Mg(s) + O_2(g) rarr 2MgO(s)`

Explanation:

`"Moles of metal"=(5.0*g)/(24.3*g*mol^-1)=0.206*mol`

`"Moles of dioxygen"=(4.0*g)/(32.0*g*mol^-1)=0.125*mol`

`"Dioxygen gas"` is in stoichiometric excess, and only `0.103` mol of dioxygen will react. And thus `0.022*mol` dioxygen will remain unreacted, i.e. approx. `0.7*g`.