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ChemistryStoichiometryBasic Methods of Calculations (POAC, LR)Medium2 minQB
ChemistryMediummatching list

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:
D
Solution:

(P)

Initial mole: 2 2 0 0

Final mole: 0 1 1

Excess reagent left:

Volume of H₂ = 22.4 lit.

Solid product obtained = 1 mole.

Limiting reagent is HCl.

(Q)

Initial mole: 0 0

Final mole: 0

Excess reagent:

Volume of gas = 11.2 lit.

Solid product = mole.

Limiting reagent is HCl.

(R)

Initial mole: 0 0

Final mole: 0 1 1

Excess reagent not present.

Volume of gas = 22.4 lit. at STP.

Solid product is 1 mole.

(S)

Initial mole: 0 0

Final mole: 0 2

No excess reagent left.

Volume of gas = 44.8 lit.

Solid product is mole.

Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Basic Methods of Calculations (POAC, LR)
2mℹ️ Source: QB

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