Limiting Reagent or Reactant
Trending Questions
Q. What is the molarity of a 4.9% H3PO4 solution by mass? (Density of H3PO4 solution = 1.22 g/mL)?
- 0.61 M
- 0.05 M
- 1.22 M
- 1.54 M
Q. 0.5 moles of H2SO4 is mixed with 0.2 moles of Ca(OH)2. The maximum number of moles of CaSO4 formed is:
Reaction: H2SO4+Ca(OH)2→CaSO4+2H2O
Reaction: H2SO4+Ca(OH)2→CaSO4+2H2O
- 0.2 mol
- 0.5 mol
- 0.4 mol
- 1.5 mol
Q. How many moles of Zn[FeS2]2 can be made from 2 g of Zn, 3 g of Fe and 4 g of S?
Zn+2Fe+4S→Zn[FeS2]2
(Given: Molar mass of Fe = 56 g/mol and molar mass of Zn = 65 g/mol)
Zn+2Fe+4S→Zn[FeS2]2
(Given: Molar mass of Fe = 56 g/mol and molar mass of Zn = 65 g/mol)
- 0.0157 mol
- 0.0265 mol
- 0.2650 mol
- 0.0538 mol
Q.
of (At Mass 24) will produce equal to?
Q. At STP, the volume of ammonia formed using 1 mole of nitrogen and 4 moles of hydrogen on complete reaction is:
- 67.2 L
- 11.2 L
- 22.4 L
- 44.8 L
Q. Consider the following balanced reaction:
2Ca(s)+O2(g)→2CaO
If 60 g of Ca is placed in a reaction chamber with 32 g of O2, which will be the limiting reactant and mass of the excess reactant left?
2Ca(s)+O2(g)→2CaO
If 60 g of Ca is placed in a reaction chamber with 32 g of O2, which will be the limiting reactant and mass of the excess reactant left?
- Calcium; 8 g
- Oxygen; 8 g
- Oxygen; 16 g
- Calcium; 10 g
Q. 10 g of Hydrogen and 64 g of Oxygen were filled in a steel vessel and exploded. Calculate the amount of water produced in the reaction.
- 5 mol
- 3 mol
- 2 mol
- 4 mol
Q. Consider the following reaction:
8Fe+S8→8FeS
If we begin with 293 g of Fe and 17.2 g of S8, how many grams of FeS will be produced?
8Fe+S8→8FeS
If we begin with 293 g of Fe and 17.2 g of S8, how many grams of FeS will be produced?
- 42.30 g
- 45.00 g
- 52.03 g
- 47.17 g
Q. How many moles of H2 will be needed to produce 6 moles of H2O when it reacts with excess of O2 ?
- 3
- 4
- 6
- 5
Q. If 27 g of carbon is mixed with 88 g of oxygen and is allowed to burn to produce CO2, then:
- Oxygen is the limiting reagent
- Volume of CO2 gas produced at NTP is 50.4 L
- C and O combine in mass ratio 3:8
- Volume of unreacted O2 at STP is 11.2 L
Q. For the reaction,
2H2(g)+O2(g)→2H2O(g)
Initially, 3 moles of H2 and 2 moles of O2 are present. Identify the correct statement.
2H2(g)+O2(g)→2H2O(g)
Initially, 3 moles of H2 and 2 moles of O2 are present. Identify the correct statement.
- H2 is the limiting reagent
- O2 is the limiting reagent
- 1 mol of H2 is left unreacted
- 1 mol of O2 is left unreacted
Q. How many grams of titanium (IV) oxide can be produced when 80 g of TiCl4 is made to react with 20 g of O2 as shown in the following equation? (Molar mass of titanium = 48 g/mol)
TiCl4(s)+O2(g)→TiO2(s)+2Cl2(g)
TiCl4(s)+O2(g)→TiO2(s)+2Cl2(g)
- 130 g
- 34 g
- 50 g
- 88 g
Q. Calculate the number of moles of Fe2O3 when 600 g of FeS2, reacts with 800 g of O2.
4FeS2+11O2→2Fe2O3+8SO2
(Given: Molar mass of Fe is 56 g/mol)
4FeS2+11O2→2Fe2O3+8SO2
(Given: Molar mass of Fe is 56 g/mol)
- 1 mol
- 1.5 mol
- 2 mol
- 2.5 mol
Q.
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
0.5
13
0.125
Q. The ratio of amounts of H2S needed to precipitate all the metal ions from 100 ml of 1 M AgNO3 and 100 ml of 1 M CuSO4 will be
- 1 : 1
- 1 : 2
- 2 : 1
- None of these
Q. Law of multiple proportions is not applicable for the oxide(s) of:
- Aluminium
- Iron
- Nitrogen
- Carbon
Q. Sodium azide (NaN3) is an explosive chemical used in automobile airbags. It is made by the following reaction:
NaNO3+3NaNH2→NaN3+3NaOH+NH3
One takes 17.0 g of NaNO3 and 13.0 g of NaNH2. If this amount is used in automobile bags, how much N2 is formed due to collision at STP?
NaNO3+3NaNH2→NaN3+3NaOH+NH3
One takes 17.0 g of NaNO3 and 13.0 g of NaNH2. If this amount is used in automobile bags, how much N2 is formed due to collision at STP?
- 3.73 L
- 12.93 L
- 8.62 L
- 2.15 L
Q. Find the moles of HCl gas obtained at STP from the following set of reactions:
FeCl3+O2→Fe2O3+Cl2(g)H2(g)+Cl2(g)→2HCl(g)
Starting with 4.56 mol of FeCl3 , 33.6 L of O2 at STP and 7 mol of H2 gas.
(Molar mass of Fe = 56 g/mol)
FeCl3+O2→Fe2O3+Cl2(g)H2(g)+Cl2(g)→2HCl(g)
Starting with 4.56 mol of FeCl3 , 33.6 L of O2 at STP and 7 mol of H2 gas.
(Molar mass of Fe = 56 g/mol)
- 2 mol
- 6 mol
- 4 mol
- 8 mol
Q. Consider the following balanced reaction:
2Ca(s)+O2(g)→2CaO
If 60 g of Ca is placed in a reaction chamber with 32 g of O2, which will be the limiting reactant and mass of the excess reactant left?
2Ca(s)+O2(g)→2CaO
If 60 g of Ca is placed in a reaction chamber with 32 g of O2, which will be the limiting reactant and mass of the excess reactant left?
- Calcium; 8 g
- Oxygen; 16 g
- Calcium; 10 g
- Oxygen; 8 g
Q. Assertion: When 4 moles of H2 react with 2 moles of O2, then 4 moles of water is formed.
Reason: O2 will act as the limiting reagent
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 true, but the reason is not the correct explanation of the assertion.
- The assertion is true, but the reason is false.
- Both the assertion and the reason are false statements.
Q. 0.006 moles of Ca(NO3)2 are added to an aqueous solution containing 0.003 moles of Na2C2O4 and the reaction is allowed to go to completion. Select the incorrect statement from the following:
- 0.003 mole of calcium oxalate will get precipitated.
- 0.001 moles of Ca2+ will remain in excess.
- Na2C2O4 is the limiting reagent.
- Ca(NO3)2 is the excess reagent.
Q. 6 moles of Na2CO3 is reacted with 10 moles of HCl. Find the volume of CO2 gas produced at STP.
The reaction is:
Na2CO3+2HCl→2NaCl+CO2+H2O
The reaction is:
Na2CO3+2HCl→2NaCl+CO2+H2O
134.4 L
- 112 L
- 22.4 L
- 44.8 L
Q.
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
Q. If 200 ml of 0.3 M H3PO3 is mixed with 200 ml of 0.3 M Ba(OH)2, find the incorrect option from the following:
- H3PO3 is limiting reagent if the reaction goes to completion
- Reactants get used up if Ba(HPO3) is the product
- H3PO3 is the limiting reagent if Ba(H2PO3)2 is the product
- Reactants get used up if Ba(OH)(H2PO3) is the product.
Q.
Starting with 6 mol of A, 9 mol of B, 4 mol of E and 15 mol of H, the number of moles of G formed will be
Reaction | Yield of the reaction |
(i) 2A+3B→3C+D | 80% |
(ii) 2C+E→4F | 40 % |
(iii) 3F+7H→8G | 50% |
Starting with 6 mol of A, 9 mol of B, 4 mol of E and 15 mol of H, the number of moles of G formed will be
- 4.48 mol
- 7.68 mol
- 5.28 mol
- 6.68 mol
Q. Suppose 316.0 g of aluminium sulphide reacts with 493.0 g of water, what mass of the excess reactant will remain?
Given reaction: Al2S3+6H2O→2Al(OH)3+3H2S
Given reaction: Al2S3+6H2O→2Al(OH)3+3H2S
- 265.14 g
- 108.52 g
- 400 g
- 66.25 g
Q. How many moles of oxygen is required to burn one mole of methane, completely ?
- 1
- 3
- 2
- 4
Q. Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major industrial chemicals and is widely used. Nitric acid is manufactured by the catalytic oxidation of ammonia that is known as Ostwald Process which can be represented by sequence of reactions shown below
4NH3+5O2(g)Catalyst−−−−−→Pt/Rh4NO(g)+6H2O(g) ...(i)
2NO(g)+O2(g)1120 K−−−−→2NO2(g) ...(ii)
3NO2(g)+H2O(l)→2HNO3(aq)+NO(g) ...(iii)
The aqueous nitric acid obtained by this method can be concenteated by distillation to 68.5% by weight. Further concentration to 98% acid can be achieved by dehydration with concentrated sulphuric acid.
85 kg of NH3 (g) was heated with 320 kg oxygen in the first step and HNO3 is prepared according to the above reactions. If the final solution has volume 500 L, then molarity of HNO3 is:
4NH3+5O2(g)Catalyst−−−−−→Pt/Rh4NO(g)+6H2O(g) ...(i)
2NO(g)+O2(g)1120 K−−−−→2NO2(g) ...(ii)
3NO2(g)+H2O(l)→2HNO3(aq)+NO(g) ...(iii)
The aqueous nitric acid obtained by this method can be concenteated by distillation to 68.5% by weight. Further concentration to 98% acid can be achieved by dehydration with concentrated sulphuric acid.
85 kg of NH3 (g) was heated with 320 kg oxygen in the first step and HNO3 is prepared according to the above reactions. If the final solution has volume 500 L, then molarity of HNO3 is:
- 2 M
- 8 M
- 3.33 M
- 6.66 M
Q. 50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g). Calculate the mass of NH3(g) formed:
- 40.0 kg
- 14.0 kg
- 25.0 kg
- 56.6 kg
Q. A reaction container holds 5.77 g of P4 and 5.77 g of O2 and the following reaction occurs: P4+O2→P4O6. The amount of excess reagent left is?
- 10.51 g
- 2.84 g
- 20.32 g
- 1.31 g