A chemist prepares a solution of ironIII chloride FeCl3 by measuring out 72.7mg of FeCl3 into a 100.mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl− anions in the chemist's solution.

The molarity of chloride anions in the solution is 0.01344 M

Explanation:

To calculate the molarity of solution, we use the equation:

Given mass of ferric chloride = 72.7 mg = 0.0727 g    (Conversion factor:  1 g = 1000 mg)

Molar mass of ferric chloride = 162.2 g/mol

Volume of solution = 100 mL

Putting values in above equation, we get:

1 mole of ferric chloride contains 1 mole of ions and 3 moles of ions

Moles of = (3 × 0.00448) = 0.01344 M

Hence, the molarity of chloride anions in the solution is 0.01344 M

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.