How much magnesium metal is needed to prepare a #25*g# mass of magnesium oxide?

1 Answer
anor277
Apr 13, 2017

Approx. #15*g# of magnesium metal.........

Explanation:

We need (i) a stoichiometric equation:

#Mg(s) + O_2(g) rarr MgO(s)#,

and (ii), equivalent quantities of the product #MgO#:

#"Moles of magnesium oxide"=(25*g)/(40.30*g*mol^-1)=0.620*mol#.

Now from the stoichiometry of the equation, if there are #0.620*mol# #MgO#, then there MUST have been at least #0.620*mol# #"magnesium metal"#, and this quantity represents a MASS of #0.620*molxx24.31*g*mol^-1=??g#

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