Calculate the change in heat (Q) that occurs when 38.68grams of steam at 304°C cools to liquid water at 197°C. (cp = 4.184 J/g°C)

Change in heat (Q) = -1.7316 x 10⁴

Explanation:

Given:

Mass m = 38.68 g

Change in temperature Δt = (197-304) = -107°C

Specific heat C = 4.184 J/g°C

Find:

Change in heat (Q)

Computation:

Change in heat (Q) = (C)(m)(Δt)

Change in heat (Q) = (4.184)(38.68)(-107)

Change in heat (Q) = -1.7316 x 10⁴

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.