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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