Chemistry Reference
In-Depth Information
25. [41.5 g]—
2Mg + O
2
2MgO
25.0 g
28.0 g
x g
mass
molar mass
25.0 g
24.3 g/
m
ole
= 1.03 moles (limiting)
A.
# of moles Mg =
=
mass
molar mass
28.0 g
32.0 g/mole
B.
# of moles O
2
=
=
= 0.875 moles
C. The magnesium is the limiting reactant because you would need 2(0.875) = 1.75
moles of magnesium to react completely with the 0.875 moles of oxygen. We
have less than that.
D. # of moles of MgO can be determined from molar ratio:
Mg
MgO
coefficients
2
1.03
2
x
# of moles
=
# of moles of MgO = 1.03 moles
40.3 g
mole
E. Mass of MgO = # of moles × molar mass = 1.03 moles ×
= 41.5 g
