How much potassium chloride is needed to make 0.500 m solution with 1.50 L of water?

55.9 g KCl.

Explanation:

Hello there!

In this case, according to the definition of molality for the 0.500-molar solution, we need to divide the moles of solute (potassium chloride) over the kilograms of solvent as shown below:

Thus, solving for the moles of solute, we obtain:

Since the density of water is 1 kg/L, we obtain the following moles:

Next, since the molar mass of KCl is 74.5513 g/mol, the mass would be:

Regards!

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.

idk i’m struggling too