Which of the following aqueous solutions contains the greatest number of ions?

a. 400. ml of 0.10 m nacl;
b. 300. ml of 0.10 m cacl2;
c. 200. ml of 0.10 m fecl3

Answer : The correct option is, (b) 300. mL of 0.10 M CaCl₂

Explanation :

We have to calculate the number of moles of ions for the following aqueous solutions.

(a)  400. mL of 0.10 M NaCl

NaCl dissociates to give 2 moles of ions.

Volume of solution = 400. mL = 0.400 L

Total moles of ion = 0.04 mol × 2 = 0.08 mol

(b)  300. mL of 0.10 M CaCl₂

CaCl₂ dissociates to give 3 moles of ions.

Volume of solution = 300. mL = 0.300 L

Total moles of ion = 0.03 mol × 3 = 0.09 mol

(c)  200. mL of 0.10 M FeCl₃

FeCl₃ dissociates to give 4 moles of ions.

Volume of solution = 200. mL = 0.200 L

Total moles of ion = 0.02 mol × 4 = 0.08 mol

From this we conclude that, aqueous solutions contains the greatest number of ions.

Hence, the correct option is, (b) 300. mL of 0.10 M CaCl₂

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.

fe 55.85 g/mol and h2o 18.02 g/mol.