Limiting Reagent or Reactant
Trending Questions
Zn+H2SO4→ZnSO4+H2
Under similar conditions of pressure and temperature, 40 ml of slightly moist hydrogen chloride gas is mixed with 20 ml of ammonia gas, the final volume of gas at the same temperature and pressure will be
- 100 ml
- 20 ml
- 40 ml
- 60 ml
- 3 moles
- 4 moles
- 1 moles
- 2 moles
The number of moles of Fe2O3 formed when 0.5 moles of O2 and 0.5 moles of Fe are allowed to react are
0.25
1/3
0.125
0.5
Which law of chemical combinations will govern your answer?
A+B⟶A3B2(unbalanced)A3B2+C⟶A3B2C2(unbalanced)
Above two reactions are carried out by taking 3 moles each of A and B and one mole of C. Then which option is wrong
1 mole of A3B2C2 is formed
1/2 mole of A3B2C2 is formed
1 mole of A3B2 is formed
1/2 mole of A3B2 is left finally
2NaOH+H2SO4→Na2SO4+2H2O
- 2
- 1
- 3
- 4
- 12
- 20
- 40
- 14.8
If 0.5 mole of BaCl2 is mixed with 0.2 mole of Na3PO4, the maximum number of moles of Ba3(PO4)2 formed are
0.5
0.3
0.7
0.1
- 10−2 mole
- 0.1 mole
- 0.5×10−2 moles
- 0.2 moles
- 1 : 1
- 1 : 2
- 2 : 1
- None of these
- 14.35 g
- 15 g
- 18 g
- 19 g
18g of steam is passed over 2g of Iron.H2O + Fe → H2 + Fe2O3 Which is the excessive reagent? [Fe = 55.85g]
- 22 g of CO2 gas
- 11.2 L of CO2 gas
- 0.1 mole of CO2 gas
- 22.4×103 mL of CO2 gas
- Law of constant proportion
- Law of conservation of mass
- Law of Multiple proportion
- None of these
In each of the following questions, an Assertion (A) is followed by a corresponding Reason (R). Use the following keys to choose the appropriate answer.
Assertion (A): If 30 mL of H2 and 20 mL of O2 react to form water, 5 mL of H2 is left at the end of the reaction.
Reason (R): H2 is the limiting reagent.
If both (A) and (R) are correct and (R) is the correct explanation for (A).
If both (A) and (R) are correct and (R) is not the correct explanation for (A).
If (A) is correct but (R) is incorrect.
If (A) is incorrect and (R) is correct.
If both (A) and (R) are incorrect.
- 0
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- True
- False
S(s)+O2(g)→SO2(g)
- 1
- 4
- 8
- 16
- 32
2VO + 3Fe2O3 → 6FeO + V2O5 If we start with 2g of VO and 5.75g of Fe2O3, which is the limiting reagent?
[V = 47.87g] [Fe = 55.85g]
VO
- (i) 25 c.c. (ii) 50 c.c.
- (i) 50 c.c. (ii) 75 c.c.
- (i) 25 c.c. (ii) 75 c.c.
- (i) 75 c.c. (ii) 75 c.c.
30 mL of hydrogen and 20 mL of oxygen react to give water. What is left at the end of the reaction?
5 mL of oxygen
10 mL of hydrogen
5 mL of hydrogen
10 mL of oxygen
Is 400g of oxygen gas enough to burn 20 moles of methane completely?
Yes
No
All the above
Insufficient data
16.8 litre gas containing H2 and O2 is formed at NTP on electrolysis of water. What should be the weight of electrolysed water
- 5 g
- 9 g
- 10 g
- 12 g
Cl2+S2O32−+OH−⟶SO42−+Cl−+H2O
Starting with 0.15 mole Cl2, 0.010 moleS2O32−and 0.30 mole OH−, mole of Cl2 left in solution will be
- 0.09
- 0.11
- 0.01
- 0.04
Reason: O2 will act as the limiting reagent
- Both the assertion and the reason are true, and the reason is the correct explanation of the assertion.
- Both the assertion and the reason are false statements.
- The assertion is true, but the reason is false.
- Both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
End point in the titration of CuSO4 iodometrically (with starch as the indicator) is the disappearance of blue color.