23 grams of sodium reacts with 244 cm3 of water that is initially at 293 k. it produces an enthalpy change of 71 kj. what is the final temperature of the water? the specific heat capacity of water is 4.18 j/k g.

Given mass of Na is 23 g

Volume of water =

Mass of water =

Total solution mass = 23 g + 244 g = 267 g

Specific heat capacity of water = 4.18 J/K.g

Equation relating mass, heat, specific heat capacity and temperature change is:

ΔT

Tfinal = 356.6 K

Therefore, the final temperature of water will be 356.6 K

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.

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: