Chemistry Reference
In-Depth Information
5.
[10 moles]—
oxygen
water
coefficients
moles
5
x
4
8
=
4x = 40, x = 10
hydrogen nitrogen
6.
[3 moles]—
coefficients
3
9
1
x
=
moles
3x = 9, x = 3
7.
[15 moles]—
hydrogen nitrogen
coefficients
moles
3
x
1
5
=
x = 15
8.
[1.5 moles]—
hydrogen ammonia
coefficients
moles
3
x
2
1
=
2x = 3, x = 1.5
nitrogen
ammonia
9.
[5 moles]—
coefficients
1
x
2
10
=
moles
2x = 10, x = 5
10. [4 moles]—You must determine the “limiting reactant,” which means that
you must figure out how much ammonia you could generate with each of
the elements, if you had plenty of the other, and figure out which reactant
allows you to produce the least.
hydrogen ammonia
nitrogen
ammonia
coefficients
moles
3
9
2
x
coefficients
moles
1
2
2
x
=
=
or x = 6 or x = 4
The hydrogen would allow us to make 6 moles of ammonia, but we only
have enough nitrogen to make 4 moles of ammonia. The nitrogen would be
called our limiting reactant, because it is limiting the amount of ammonia
that we can make to four moles. The hydrogen would be called our excess
reactant, because we have more than we can actually use.
Lesson 7-4 Review
1.
[1.51 g of Mg]—
2Mg
(s)
+ O
2(g)
2MgO
(s)
x g
2.50 g
mass
molar mass
2.50 g
40.3 g/mole
A.
# of moles of MgO =
=
= 0.0620 moles
B. # of moles of Mg can be found with the molar ratio:
Mg
MgO
coefficients
moles
3
9
2
x
=
# of moles of Mg (x) = 0.0620 moles
C.
Mass of Mg = # of moles × molar mass = 0.0620 moles × 24.3 g/mole =1.51 g of Mg
