Tutorial 1 - answers
• If 20.0 mL of a 0.100 M solution of sodium phosphate is mixed with 25.0 mL of a
0.200 M solution of zinc chloride, what mass of zinc phosphate will precipitate from
the reaction?
Marks
6
Steps: 1) Write equation for the reaction; 2) Determine amount (in moles) of starting
materials present; 3) Determine which of the starting materials is limiting; 4) Based on
the limiting reagent, determine the amount (in moles) of product obtained; 5) Use molar
mass of product to determine mass of product obtained.
1. 3Zn
2+
(aq) + 2PO
4 3-
(aq) → Zn
3
(PO
4
)
2
(s)
2. 25.0 mL of a 0.200 M solution of ZnCl
2
contains:
n(Zn
2+
) = concentration × volume = (0.200 mol L
-1
) × (25.0 × 10
-3
L)
= 0.00500 mol
20.0 mL of a 0.100 solution of Na
3
PO
4
contains:
n(PO
43-
) = (0.100 mol L
-1
) × (20.0 × 10
-3
L) = 0.00200 mol
3. Determine which reagent is limiting: if Zn
2+
is limiting it would react with
(2/3)0.00500 = 0.00333 mol of phosphate – the amount of phosphate present is less
than this therefore phosphate is limiting. Put another way, if phosphate was
limiting it would react with (3/2)0.00200 = 0.00300 mol of zinc – the amount of zinc
present is more than this therefore phosphate is limiting.
4. Based on the amount (moles) of phosphate present, the maximum amount of
product formed is: ½ × 0.00200 mol = 0.00100 mol
5. The molar mass of zinc phosphate is:
(3 × 65.39 (Zn)) + 2 × (30.97 (P) + 4 × 16.00 (O)) = 386.11g mol
-1-
The mass of zinc phosphate produced is therefore:
mass = number of moles × molar mass = 0.00100 × 386.11 = 0.386 g
Answer: 0.386 g
What is the final concentration of zinc ions in solution after the above reaction?
The number of moles of Zn
2+
removed by precipitation: 3 × 0.00100 = 0.00300 mol
The amount remaining is therefore: 0.00500 – 0.00300 = 0.00200 mol
The total volume of the solution after mixing is (20.0 + 25.0) = 45.0 mL so the
concentration is:
[Zn
2+
] = number of moles/volume in L
= 0.00200 mol /(45.0 × 10
-3
L) = 0.0444 M
Answer: 0.0444 M
