**Solutions
to p 17** (green
book)

I've done some problems by proportion for those of you who prefer that method and I've also done some the faster way, like I often do it in class.

1. a.

NH_{3 }O_{2}

*x *= 2.50 moles of O_{2 }

_{}_{}

b.

O_{2 }NO

_{}_{}

*x* = 8.00 moles of NO

_{}

_{}

c. NH_{3 }O_{2}

*x* = 2.5 moles
O_{2}

2. 47.3 ml = 0.0473 L of H_{2}, so at STP (see
conditions in problem)

_{}_{}

(this step is equivalent to a ratio used in previous solutions)

3.

At 0^{o}C and 101.3 kPa, we would have produced
3.0 X 22.4 L/mole = 67 L, but at

40^{o}C and 101.3 kPa(we assume), we will produce
more. According to Charles' Law,

V_{2} = 77 L

4. 1.204
X 10^{24} molecules. Divide by Avogadro's number and you realize that
we have 2.00 moles of Ne in container B.

Container A has 4 moles of Ar under the same conditions. With twice the number of moles, A must be twice as big in volume as B.

5. a. Consider 1.00 moles of Ar = 39.9 g and 22.4 L at STP, so its STP density is 39.9 g/22.4 L = 1.78 g/L

b. At
-50^{o}C, using Charles' Law the volume of 1.00 moles of Ar is reduced
to 18.3 L, so its

density increases to 39.9 g/18.3 L = 2.18 g/L

6. from the board:

Question was:

Given, 3 H_{2} + N_{2} à2 NH_{3}

If 11.2 L of H_{2} and 11.2
L of N_{2} are mixed, what’s the most NH_{3} that can be
produced? Which gas will be in excess, and how many liters will be in excess?

__Solution:
__

__Longer Method
__

11.2 L of an ideal gas is 0.500 moles at STP

So we have 0.500 moles of hydrogen and 0.500 moles of nitrogen.

3 H_{2} + N_{2} à2 NH_{3}

If you assume that 0.500 moles of N_{2} react, then
according to the ratio , you would need 3*0.500 =1.500 moles of hydrogen. But
we only have 0.500, so the N_{2} can’t all be reacting.

If we assume that all 0.500 moles of H_{2} react,
then only 0.500/3 moles of N_{2} react, which means that there will be
0.500 – 0.500/3 leftover moles of N_{2} = 0.333..moles = 7.47 L
leftover nitrogen.

Since all 0.500 moles of H_{2} react, according to
the ratio, 2/3(0.500) moles of NH_{3} are produced = 0.333..moles =
7.47 L of NH_{3.
}

__Shorter Method(L of ideal gases are proportional to moles
at same T,P (avogadro’s Hypothesis))
__

We have 11.2 L of hydrogen and 11.2 L of nitrogen.

3 H_{2} + N_{2} à2 NH_{3}

If you assume that 11.2 L of nitrogen of N_{2} react, then according to the ratio , you
would need 3*11.2 L =33.6 L of hydrogen. But we only have 11.2 L, so the N_{2} can’t all be reacting.

If we assume that all 11.2 L of H_{2} react, then
only 11.2 L/3 moles of N_{2} react, which means that there will be 11.2
– 11.2/3= 7.47 L leftover nitrogen.

Since all 11.2 L of H_{2} react, according to the
ratio, 2/3(11.2 L) = 7.47 L of NH_{3.
}