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. 

            NH₃                                       O₂

x = 2.50 moles of  O₂

b.

            O₂                          NO

x = 8.00 moles of NO

c.          NH₃                       O₂

x = 2.5 moles O₂

2.                  47.3 ml = 0.0473 L of H₂, so at STP (see conditions in problem)

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

3.        

           

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

            40^oC and 101.3 kPa(we assume), we will produce more. According to Charles' Law,

V₂ = 77 L

4.         1.204 X 10²⁴ 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^oC, 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 H₂ + N₂ à2 NH₃

If 11.2 L of H₂ and 11.2 L of N₂ are mixed, what’s the most NH₃ 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 H₂ + N₂ à2 NH₃

If you assume that 0.500 moles of N₂ 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 N₂ can’t all be reacting.

If we assume that all 0.500 moles of H₂ react, then only 0.500/3 moles of N₂ react, which means that there will be 0.500 – 0.500/3 leftover moles of N₂ = 0.333..moles = 7.47 L leftover nitrogen.

Since all 0.500 moles of H₂ react, according to the ratio, 2/3(0.500) moles of NH₃ 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 H₂ + N₂ à2 NH₃

If you assume that 11.2 L  of nitrogen of N₂ 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 N₂ can’t all be reacting.

If we assume that all 11.2 L of H₂ react, then only 11.2 L/3 moles of N₂ react, which means that there will be 11.2 – 11.2/3= 7.47 L leftover nitrogen.

Since all 11.2 L of H₂ react, according to the ratio, 2/3(11.2 L)  = 7.47 L of NH_3.