A #3.0*g# mass of magnesium metal is combusted in a #2.0*g# mass of oxygen gas. What mass of magnesium oxide results?

1 Answer
Jan 21, 2018

Approx. #5.0*g#

Explanation:

We must look at the stoichiometric equivalence, for which we need a stoichiometric equation...

#Mg(s) + 1/2O_2(g) rarr MgO(s)#

And we interrogate the molar quantities of each reagent in the first scenario...

#"Moles of magnesium metal"=(3.0*g)/(24.3*g*mol^-1)=0.0124*mol#

#"Moles of dioxygen gas"=(2.0*g)/(32.0*g*mol^-1)=0.0625*mol#

There is thus stoichiometric dioxygen gas in this instance....and we get #5.0*g# of product out....

For the second scenario.....

#"Moles of magnesium metal"=(3.0*g)/(24.3*g*mol^-1)=0.0124*mol#

#"Moles of dioxygen gas"=(5.04*g)/(32.0*g*mol^-1)=0.158*mol#

Here dioxygen is in stoichiometric excess....we still get out #5.0*g# of product....but a #3.04*g# mass of dioxygen gas remains unreacted.. Do you follow?