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Question 6
This question is about oxygen (O₂) and sulfur dioxide (SO₂). 1. Give the test and result for oxygen gas. Test: ________________________ Result: __________________... show full transcript
Step 1
Step 2
Answer
When some of the sulfur trioxide (SO₃) is removed, the concentration of SO₃ decreases. According to Le Chatelier's principle, the equilibrium will shift to the right-hand side to favor the production of more SO₃, thus attempting to re-establish equilibrium.
Step 3
Answer
First, calculate the number of moles of calcium oxide (CaO):
Molar mass of CaO = 40 (Ca) + 16 (O) = 56 g/mol.
Number of moles of CaO = ( \frac{7.00 \text{ g}}{56 \text{ g/mol}} = 0.125 \text{ moles} ).
From the reaction, 1 mole of CaO produces 1 mole of CaSO₃. Therefore, 0.125 moles of CaO will produce 0.125 moles of CaSO₃.
Now, calculate the mass of calcium sulfite (CaSO₃):
Molar mass of CaSO₃ = 40 (Ca) + 32 (S) + 48 (O) = 120 g/mol.
Mass of CaSO₃ = ( 0.125 \text{ moles} \times 120 \text{ g/mol} = 15.0 \text{ g} ).
Thus, the mass of calcium sulfite produced is 15.0 g.
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