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

1 Answer
anor277
Feb 21, 2017

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.

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