How many grams of water can be heated from 5.0∘c to 37.0∘c with 20.0j of heat? the specific heat of water is 4.184j/g∘c.

-  mass of water is 0.15 g.

Solution:- This problem is based on the formula:

where, q is the heat energy, m is mass, is specific heat and is change in temperature.

 = final temperature - initial temperature

 = 37.0 - 5.0 = 32.0 degree C

m = ?

q = 20.0 J

Let's plug in the values in the formula:

m = 0.15 g

So, the mass of water for the given problem is 0.15 g.

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.

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