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In an experiment, 172.1 g of gypsum, CaSO₄·2H₂O (M = 172.1 g mol⁻¹), was heated to constant mass in a large crucible - VCE - SSCE Chemistry - Question 13 - 2011 - Paper 1
Question 13
In an experiment, 172.1 g of gypsum, CaSO₄·2H₂O (M = 172.1 g mol⁻¹), was heated to constant mass in a large crucible. The loss in mass of the crucible and contents w... show full transcript
Worked Solution & Example Answer:In an experiment, 172.1 g of gypsum, CaSO₄·2H₂O (M = 172.1 g mol⁻¹), was heated to constant mass in a large crucible - VCE - SSCE Chemistry - Question 13 - 2011 - Paper 1
Step 1
Identify the Reaction
Answer
When gypsum (CaSO₄·2H₂O) is heated, it undergoes a dehydration reaction, releasing water (H₂O) vapor. In this case, the correct representation of the reaction at elevated temperatures is:
ext{CaSO}_4 ullet 2 ext{H}_2 ext{O}(s) ightarrow ext{CaSO}_4(s) + 2 ext{H}_2 ext{O}(g)Step 2
Calculate the Mass Loss
Answer
The mass loss observed in the experiment was given as 27.0 g. This mass directly correlates to the water released when gypsum is heated. The reaction equation shows that 2 moles of water are produced for every mole of gypsum.
Given:
- Molar mass of CaSO₄·2H₂O = 172.1 g/mol
- Mass loss = 27.0 g
To find the moles of water lost:
ext{Moles of water} = rac{ ext{mass lost}}{ ext{Molar mass of water}} = rac{27.0 g}{18.02 g/mol} \ ext{ approximately equals } 1.5 ext{ moles}Step 3