Chemistry - Stoichiometry Question with Solution | TestHub
Match List-I correctly with List-II :
List - I | List - II |
|---|---|
(P) Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2(g) above reaction is carried out by taking 2 moles each of Zn and HCl | (1) 50% of excess reagent left |
(Q) above reaction is carried out by taking 170 g and 18.25 g HCl ( ) | (2) 22.7 L of gas at STP is liberated |
(R) CaCO3 (s) → CaO(s) + CO2 (g) 100 g CaCO3 is decomposed | (3) 1 moles of solid product is obtained |
(S) 2KClO3 (s) → 2KCl(s) + 3O2 (g) 2/3 moles of KClO3 decomposed | (4) HCl is the limiting reagent |
| (5) None of the reagent is limiting |
Options:
Answer:
Solution:
P: Zn + 2HCl -> ZnCl₂ + H₂. Initial moles: Zn=2, HCl=2. HCl is limiting. Moles of Zn reacted = 1. Moles of Zn left = 1. % excess Zn left = . H₂ produced = 1 mole = 22.7 L at STP. HCl is limiting reagent.
Matches: (1), (2), (4)
Q: AgNO₃ + HCl -> AgCl + HNO₃. Molar mass AgNO₃ = 170 g/mol, HCl = 36.5 g/mol. Initial moles: AgNO₃ = , HCl = . HCl is limiting. AgCl produced = 0.5 moles. HNO₃ produced = 0.5 moles. AgNO₃ left = 0.5 moles. % excess AgNO₃ left = . No gas liberated.
Matches: (1), (4)
R: CaCO₃ -> CaO + CO₂. Molar mass CaCO₃ = 100 g/mol. Initial moles: CaCO₃ = . CaO produced = 1 mole (solid product). CO₂ produced = 1 mole = 22.7 L at STP. No limiting reagent.
Matches: (2), (3)
S: 2KClO₃ -> 2KCl + 3O₂. Initial moles: KClO₃ = . KCl produced = moles. O₂ produced = mole = 22.7 L at STP. No limiting reagent.
Matches: (2)
Final Matching:
P -> (1), (2), (4)
Q -> (1), (4)
R -> (2), (3)
S -> (2)
