Reversing the calculations
3 marks · EasyThe following table shows the reaction between sodium sulfate and barium chloride:
| Reactant | Volume (cm³) | Concentration (mol/dm³) |
|---|---|---|
| Sodium sulfate | 50.0 | 0.100 |
| Barium chloride | excess | — |
Use the data given to answer the following:
Calculate the number of moles of sodium sulfate used.
Use the balanced equation to state how many moles of barium sulfate will be formed.
State the minimum volume of sodium sulfate solution (in cm³) needed to ensure 0.0100 mol of barium sulfate forms, using the same concentration.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) 0.00500 mol
(b) 0.00500 mol
(c) 100 cm³
Common mistakes: Incorrect calculation of moles; Forgetting ratios
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Marking points
M1 (a) · 1 mark · Correct calculation of moles of sodium sulfate (0.100 × 50.0/1000 = 0.005 mol)
B1 (b) · 1 mark · States moles of barium sulfate formed is the same as sodium sulfate, 0.005 mol (from 1:1 ratio) (follow-through)
A1 (c) · 1 mark · Calculates the correct volume of sodium sulfate solution required (0.0100/(0.100) × 1000 = 100 cm³)
Worked solution
a) Moles of sodium sulfate = 0.100 × 50.0/1000 = 0.00500 mol
b) From the equation, 1 mol Na₂SO₄ forms 1 mol BaSO₄, so 0.00500 mol is formed.
c) To form 0.0100 mol at 0.100 mol/dm³: volume = 0.0100 / 0.100 = 0.100 dm³ = 100 cm³