Answer:
9.4 metric ton of SO₂ are emitted from yhe smoke stack each year.
Explanation:
Total amount = 100 metric ton
Sulphur in the total amount = 5% = 5 metric ton
Ash = 3% = 3 metric ton
Sulphur in the ash = 10% of 3 metric ton = 0.3 metric ton
Sulphur in the flue gases = 5 metric ton - 0.3 metric ton = 4.7 metric ton
Combustion of sulphur: S (s) + O₂ (g) → SO₂ (g)
32 g S _________ 64 g SO₂
4.7 x 10⁶ g ______ x
x = 9.4 x 10⁶ g = 9.4 metric ton of SO₂
The equation is:4Al+3O2----2Al2O3
Missing question:
A. [3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s)] / 2
<span>B. 3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s) </span>
<span>C. 26.3 kJ/1 mol Fe2O3 (s) / 3.40 mol Fe2O3 (s) </span>
<span>D. 26.3 kJ/1 mol Fe2O3 (s) – 3.40 mol Fe2O3 (s).
</span>Answer is: B.
Chemical reaction: F<span>e</span>₂O₃<span>(s) + 3CO(g) → 2Fe(s) + 3CO</span>₂<span>(g);</span>ΔH = <span>+ 26.3 kJ.
When one mole of iron(III) oxide reacts 26,3 kJ of energy is required and for 3,2 moles of iron(III) oxide 3,2 times more energy is required.</span>
