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Q.lithium oxide is used to remove water from air according to the reaction- Li 2 O(s) + H 2 O(g) – 2H 2 O + 3S 1.Which reactant is limiting 2.How maNY Kg of excess reactant is left Q.0.5 mole of each of H 2 S and SO 2 mixed together in a reacting flask, react according to equation- 2H 2 S + SO 2 - 2H 2 O + 3S 1.Calculate the number of moles of S formed 2.Calculate the volume of 0.015 M HCl solution required to prepare 250 mL of a 8.25 * 10 -5 M HCl solution

Q.lithium oxide is used to remove water from air according to the reaction-
Li2O(s) + H2O(g) – 2H2O + 3S
1.Which reactant is limiting
2.How maNY Kg of excess reactant is left
Q.0.5 mole of each of H2S and SO2 mixed together in a reacting flask, react according to equation-
2H2S + SO2 - 2H2O + 3S
1.Calculate the number of moles of S formed
2.Calculate the volume of 0.015 M HCl solution required to prepare 250 mL of a 8.25 * 10-5 M HCl solution

Grade:11

1 Answers

Kaustav Pratick Hazarika
15 Points
3 years ago
Molecular mass of Li2O = 30g/mol or 0.03kg
Molecular mass of H2O = 18g/mol or 0.018kg
Molecular mass of 2LiOH = 48/mol or 0.048kg
0.03 kg of Li2OH ---- 0.018kg H2
Therefore, 
35kg Li2O-----0.018/0.03*35
                        =18/30*35
                        =21kg of H2O
Since, ..There are 72 kg of H2O available 
Thus, H2O is excess reactant.
0.018 kg of H2O-------- 0.03kg Li2O
Therefore,
72 kg of H2O -------- 0.03/0.018*72
                                   = 120kg of Li2O
Since there is only 35 kg of Li2O is available 
Therefore, Li2O is limiting reagent
Thus, excess reagent i.e. H2O left = 72kg - 21kg
= 51kg of H2O
Since, there is only 35 kg of Li

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