When hydrochloric acid is added to aluminium sulfide, the highly toxic gas hydrogen sulfide is evolved. The equation for this reaction is $$\text{Al}_2\text{S}_3(s) + 6\text{HCl}(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{H}_2\text{S}(g)$$ If excess hydrochloric acid is added to 0.200 mol of aluminium sulfide, then the volume of hydrogen sulfide produced at standard laboratory conditions (SLC) will be

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When hydrochloric acid is added to aluminium sulfide, the highly toxic gas hydrogen sulfide is evolved - VCE - SSCE Chemistry - Question 8 - 2014 - Paper 1

Question 8

When hydrochloric acid is added to aluminium sulfide, the highly toxic gas hydrogen sulfide is evolved. The equation for this reaction is $$\text{Al}_2\text{S}_3(s)... show full transcript

Worked Solution & Example Answer:When hydrochloric acid is added to aluminium sulfide, the highly toxic gas hydrogen sulfide is evolved - VCE - SSCE Chemistry - Question 8 - 2014 - Paper 1

Step 1

Calculate moles of hydrogen sulfide produced

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Answer

From the balanced equation, 1 mole of (\text{Al}_2\text{S}_3) produces 3 moles of (\text{H}_2\text{S}). Thus, 0.200 moles of (\text{Al}_2\text{S}_3) will produce:

$0.200 \text{ mol Al}_2\text{S}_3 \times \frac{3 \text{ mol H}_2\text{S}}{1 \text{ mol Al}_2\text{S}_3} = 0.600 \text{ mol H}_2\text{S}$

Step 2

Convert moles of hydrogen sulfide to volume at SLC

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Answer

At standard laboratory conditions (SLC), 1 mole of a gas occupies 24.0 L. Therefore, the volume of hydrogen sulfide produced is:

$0.600 \text{ mol H}_2\text{S} \times 24.0 \frac{\text{L}}{\text{mol}} = 14.4 \text{ L}$

Considering the options provided, the closest answer is 14.7 L.

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