Question

A 1.005 g sample of an unknown alkaline-earth metal was allowed to react with a volume of chlorine gas that contains 1.91 × 10²² Cl₂ molecules. The resulting metal chloride was analyzed for chlorine by dissolving a 0.436 g sample in water and adding an excess of AgNO₃(aq) to give a precipitate of 1.126 g of solid AgCl.

1. What is the percent Cl in the alkaline-earth chloride?
2. What is the identity of the alkaline-earth metal?
3. Write balanced equations for all chemical reactions.
4. In the reaction of the alkaline-earth metal with chlorine, which reactant is in excess and how many grams of the reactant remain unreacted?

Answer 1

#.(a): Moles of solid AgCl formed = mass of AgCl / molar mass of AgCl = 1.126 g / 143.32 g/mol = 0.007856545 mol AgCl

Since 1 mole AgCl contains 1 mol Cl, moles of Cl in 0.436 g metal chloride sample = 0.007856545 mol Cl

=> Grams of Cl present in 0.436 g metal chloride sample

= 0.007856545 mol Cl * Atomic mass of Cl = 0.007856545 mol*35.453 g/mol

= 0.27854 g Cl

=> Percent Cl in the alkaline-earth chloride = ( 0.27854 g / 0.436 g)*100 = 63.88 % (Answer)

#.(b):

Grams of metal in 0.436 g metal chloride sample = 0.436 g - 0.27854 g = 0.157462 g M

Since this is an alkaline earth metal, the chemical formula of the metal chloride will be MCl₂.

i.e 1 mol metal atom contains 2 mol Cl atom.

Hence moles of metal in 0.436 g MCl₂ =  0.007856545 mol Cl * (1 mol M / 2 mol Cl) = 0.00392827 mol M

=> Molar mass of metal = mass/mol = 0.157462 g M / 0.00392827 mol M = 40.08 g/mol

Since 40.08 g/mol is the atomic mass of Calcium, the identity of the alkaline earth metal is CaCl₂ (Answer)

#.(c):

Balanced equations:

Ca(s) + Cl₂(g) ---> CaCl₂(s)

CaCl₂(aq) + 2AgNO₃(aq) -----> 2AgCl(s) + Ca(NO₃)₂(aq)

(d): Moles of Ca = 1.005 g / 40.08 g/mol = 0.02507485 mol

Moles of Cl = 1.91*10²² * (1 mol / 6.022*10²³) = 0.031717 mol

Since the mole ratio is 1:1 and Ca has smaller mole, Ca is the limiting reactant.

and Cl₂(g) is in excess.

Moles of Cl₂(g) is in excess = 0.031717 mol - 0.02507485 mol = 0.006642 mol Cl₂(g)

Grams remain unreacted = 0.006642 mol Cl₂(g) * 70.9 g/mol = 0.471 g (Answer)