Explanation:

Moles of hydrogen 
$=\frac{\mathrm{10/}}{2}=5\text{mol}$
Moles of oxygen 
$=\frac{\mathrm{64/}}{32}=2\text{mol}$
$2H_2+O_2\to 2H_{2}O$

Thus, $O_2$ is the limiting reactant.
$\therefore$ amount of water produced $=2\times 2\text{mol}$
$=4\text{mol}$

Audio Explanation

Explanation:

Balanced equation:
N₂(g) + 3 H₂(g) → 2 NH₃(g)

n(H₂) = (3 mol H₂ / 2 mol NH₃) × 5.0 mol NH₃
= (3/2) × 5.0 mol
= 7.5 mol H₂

Explanation:

Moles of Ca = 10 / 40
≈ 0.25 mol

Moles of H₂O = 10/ 18
≈ 0.55 mol

Limiting Reagent Trick:
Ca + 2H₂O → Ca(OH)₂ + H₂
divide the given moles with the coefficient
For Ca:
0.25/1=0.25
For H₂O:
0.55/2 ≈ 0.28
So: 0.25 is less than 0.28
so we can say Ca is the limiting reagent

From the equation, 1 mol Ca → 1 mol H₂,
so H₂ formed =  ≈ 0.25 mol.

Explanation:

If 4.4 moles of ZnCO3​ are decomposed, then 4.4 moles of CO2​ will be produced, as the mole ratio is 1:1. To find the mass:

$$4.4\text{moles}\times 44\text{g/mol}=193.6\text{g}$$

Explanation:

The reaction requires equal moles of iron and sulphur (1:1 ratio).
Given 56 g of iron (1 mole) reacts with only $\text{g}$ of sulphur (1 mole),
leaving 64−32=32 g of sulphur in excess.

Explanation:

lets write the balance chemical equation:

From the balanced equation:

• 4 moles of Al react with 3 moles of O₂.

So,

Explanation:

Use this short trick:
Given/Coefficient
(Just divide the given mole with Coefficient value the one with smaller answer is limiting reagent ) 
for O2
Given is 2 and Coefficient is 1
so 2/1=2
for H2 Given is 5 and Coefficient is 2
so 5/2=2.5
we can clearly see 2 is less than 2.5 so O2 is limiting reagent

Explanation:

its a balance chemical equation Ratio is exactly 1:3 as per the chemical equation so we do not have any limiting reagent here.
N2 + 3H2 → 2NH3.

n(N2)=28/28=1 mol;
n(H2)=6/2=3 mol.
Ratio exactly 1:3,
so neither limits; both are fully consumed.

Explanation:

Balanced reaction:
2 H2 + 1 O2 → 2 H2O
Required mole ratio
H2:O2 = 2:1.

• Limiting check trick:
divide moles by coefficients. Smaller value limits.
Here: H₂ → 5/2= 2.5;
O₂ → 2/1 = 2
⇒ O₂ is limiting.

• Leftover (excess) reactant: H₂ left = 5 − (2×2) = 1 mol.

• Product from limiter: H₂O = 2 × (O₂ used) = 2 × 2 = 4 mol.

• H2 consumed = 4 mol; H2 left over = 5 − 4 = 1 mol (excess).

• H2O formed = 2×(moles of O2 used) = 2×2 = 4 mol.

TRICK

Moles ko apne stoichiometry coefficient se divide karo; jo small ho woh limiting. 

Explanation:

The required mole ratio (N_2​:H₂​) is 1:3.

Divide moles by coefficient:

N₂​: 5.0/1=5.0
H_2​: 9.0/3=3.0

The smallest resulting value (3.0 for H2​) indicates the limiting reactant.

Explanation:

The limiting reactant dictates exactly when the reaction must stop, thus setting the upper bound for the amount of product that can theoretically be formed.

Explanation:

Step 1: Write the balanced equation

N₂ + 2H₂→2NH₃

Step 2: Calculate moles of H₂

Step 3: Use the stoichiometric ratio

From the equation:
3 mol H₂ → 2 mol NH₃

So,

Step 4: Convert moles of NH₃ to volume at STP

At STP, 1 mol of any gas = 22.4 dm³

Closest to 24dm³

Explanation:

(a) Ca + 2HCl → CaCl₂ + H₂
→ 1 mol Ca gives 1 mol H₂

(b) 2Al + 6HCl → 2AlCl₃ + 3H₂
→ 2 mol Al give 3 mol H₂
→ 1 mol Al gives 1.5 mol H₂

(c) Zn + 2HCl → ZnCl₂ + H₂
→ 1 mol Zn gives 1 mol H₂

(d) 2Na + 2HCl → 2NaCl + H₂
→ 2 mol Na give 1 mol H₂
→ 1 mol Na gives 0.5 mol H₂

Explanation:

2 SO2 + O2 → 2 SO3
Required volume (mole) ratio SO2:O2 = 2:1.
Given equal volumes ⇒ ratio 1:1.
Compared to 2:1,
SO2 is limiting; O2 is in excess.
Hence O2 remains unreacted.

Explanation:

Calculate the moles of ZnS:
Moles = Mass / Molar Mass
= 9.7 g / 97.5 g/mol
≈ 0.1 moles of ZnS.

From the equation 2ZnS ⟶ 2SO₂,
the mole ratio is 1:1, so 0.1 moles of SO₂ are produced.

Calculate the volume at Room Temperature and Pressure (RTP), where 1 mole of any gas is 24 dm³.
Volume = 0.1 mol × 24 dm³/mol
= 2.4 dm³.

Explanation:

Reaction:
2 SO₂ + O₂ → 2 SO₃
For 4 mol SO₂, needed O₂ = 1/2×4=2 mol.

Explanation:

Balanced: C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O.
For 2 moles butane: 2 × 6.5 = 13 moles O₂.

Explanation:

The limiting reactant is completely consumed first in a reaction and thus limits the amount of product formed (i.e., gives the least quantity of product).

“Stoichiometric amount” means exactly the proportions required by the balanced equation, “reactant in excess” is the one left over, and “stoichiometry” is the calculation method, not a reactant.

Explanation:

A limiting reactant is the substance that is completely consumed first in a chemical reaction.
Once it’s used up, no more product can form, thus it limits the reaction and determines the minimum (actual) amount of product possible.
Excess reactants remain unreacted after the reaction is complete.

Explanation:

Balanced equation: 4Al + 3O₂ → 2Al₂O₃.
27 g Al = 1 mol Al; required O₂ moles = (3/4)×1 = 0.75 mol.
Mass of O₂ = 0.75 × 32 g/mol = 24 g.

Explanation:

Step 1: Calculate moles of reactants

• Mass of CO₂ = 120 g
  Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol
  Moles of CO₂ = 120 g / 44 g/mol = 2.72 mol

• Mass of H₂O = 80 g
  Molar mass of H₂O = (2 × 1) + 16 = 18 g/mol
  Moles of H₂O = 80 g / 18 g/mol = 4.44 mol

Step 2: Use the mole ratio from the balanced equation

From the balanced equation:
1 mol CO₂ reacts with 1 mol H₂O
So, to react with 2.72 mol of CO₂, we need 2.72 mol of H₂O

But we have 4.44 mol of H₂O, which is more than enough.
 Limiting Reagent = CO₂, because it will be completely used first.

Explanation:

• Mass of N₂ available = 140 g

• Molar mass of N₂ = 28 g/mol

• Moles of Al = 2.5 mol

• Balanced equation:
  2Al + N₂ → 2AlN

• from balanced chemical equation we know that Aluminium is limiting reagent while N2 is excess

Step 1: Calculate moles of nitrogen (N₂) available

Moles of N₂ = Mass / Molar mass
Moles of N₂ = 140 g / 28 g/mol = 5 mol

Step 2: Determine how many moles of N₂ are required

From the balanced chemical equation:
2 mol Al reacts with 1 mol N₂
So, 2.5 mol Al would react with:
(1 mol N₂ / 2 mol Al) × 2.5 mol Al = 1.25 mol N₂

Step 3: Calculate moles of N₂ remaining

Remaining moles of N₂ = Initial − Used
Remaining moles = 5 mol − 1.25 mol = 3.75 mol

Step 4: Convert remaining moles of N₂ to mass

Mass = Moles × Molar mass
Mass = 3.75 mol × 28 g/mol = 105 g

Explanation:

Moles of Al
Molar mass of Al = 27 g/mol
Moles of Al = 67.5 g / 27 g/mol = 2.5 moles

• From the balanced equation:
  2 mol Al → 2 mol AlN
  So, 2.5 mol Al will produce 2.5 mol AlN

• Molar mass of AlN = 27 (Al) + 14 (N) = 41 g/mol

• Mass of AlN = 2.5 mol × 41 g/mol = 102.5 g

Explanation:

Step 1: Calculate Moles of Reactants

• Moles of Al
  Molar mass of Al = 27 g/mol
  Moles of Al = 67.5 g / 27 g/mol = 2.5 moles

• Moles of N₂
  Molar mass of N₂ = 2 × 14 = 28 g/mol
  Moles of N₂ = 140 g / 28 g/mol = 5 moles

Step 2: Determine Limiting Reactant

From the balanced equation:
2Al + N₂ → 2AlN

• 2 moles of Al react with 1 mole of N₂

• So, 2.5 moles of Al would need:
  (1 mol N₂ / 2 mol Al) × 2.5 mol Al = 1.25 mol of N₂

But we have 5 mol of N₂, which is more than enough.
Limiting Reactant = Aluminum (Al)

Explanation:

Moles of O₂
Molar mass = 2 × 16 = 32 g/mol
Moles = 6.4 g / 32 g/mol = 0.2 mol

NOW

Step 1: Use mole ratio from equation
3 mol O₂ → 2 mol CO₂
So, 0.2 mol O₂ →
(0.2 × 2) / 3 = 0.133 mol CO₂

Step 2: Calculate mass of CO₂
Molar mass of CO₂ = 12 + (16 × 2) = 44 g/mol
Mass = 0.133 mol × 44 g/mol = 5.867 g

Explanation:

Step 1: Calculate moles of each reactant

• Moles of C₂H₄
  Molar mass = (2 × 12) + (4 × 1) = 28 g/mol
  Moles = 2.8 g / 28 g/mol = 0.1 mol

• Moles of O₂
  Molar mass = 2 × 16 = 32 g/mol
  Moles = 6.4 g / 32 g/mol = 0.2 mol

Step 2: Use mole ratio to find the limiting reactant
From the balanced equation:
1 mol of C₂H₄ reacts with 3 mol of O₂
So, 0.1 mol of C₂H₄ would require:
0.1 × 3 = 0.3 mol of O₂

Only 0.2 mol of O₂ is available, which is less than 0.3 mol.

 Limiting Reactant = O₂

Explanation:

Calculate moles of Mg:

• Mass of Mg = 1.50 g

• Molar mass of Mg = 24 g/mol

• Moles of Mg = (Mass of Mg) / (Molar mass of Mg)
  = 1.50 g / 24 g/mol = 0.0625 moles

Calculate moles of S:

• Mass of S = 1.50 g

• Molar mass of S = 32 g/mol

• Moles of S = (Mass of S) / (Molar mass of S)
  = 1.50 g / 32 g/mol = 0.0468 moles

Determine the Limiting Reactant:

From the balanced equation:
1 mole of Mg reacts with 1 mole of S to produce 1 mole of MgS.

Since there are fewer moles of S (0.0468) than Mg (0.0625), Sulphur (S) is the limiting reactant.

Calculate moles of MgS produced:

Since S is the limiting reactant, the moles of MgS produced will be equal to the moles of S used.

• Moles of MgS = 0.0468 moles

Calculate mass of MgS produced:

• Molar mass of MgS = 24 (Mg) + 32 (S) = 56 g/mol

• Mass of MgS = Moles × Molar mass = 0.0468 × 56 = 2.6208 g

Explanation:

Step 1: Calculate moles of KClO₃

• Molar mass of KClO₃ = 39 (K) + 35.5 (Cl) + 3 × 16 (O) = 122.5 g/mol

• Moles of KClO₃ = (Given mass) / (Molar mass)
  = 49 g / 122.5 g/mol = 0.4 moles

Step 2: Determine moles of O₂ produced

• From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂

• Therefore, 1 mole of KClO₃ produces 3/2 moles of O₂

• So, 0.4 moles of KClO₃ will produce (3/2) × 0.4 = 0.6 moles of O₂

Step 3: Calculate volume of O₂ at STP

• At STP, 1 mole of any gas occupies 22.4 dm³

• Volume of 0.6 moles of O₂ = 0.6 × 22.4 = 13.44 dm³

Explanation:

Step 1: Calculate moles of CaCO₃

• Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3 × 16 (O) = 100 g/mol

• Moles of CaCO₃ = (Given mass) / (Molar mass)
  = 50 g / 100 g/mol = 0.5 moles

Step 2: Determine moles of CO₂ produced

• From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂

• Therefore, 0.5 moles of CaCO₃ will produce 0.5 moles of CO₂

Step 3: Calculate mass of CO₂

• Molar mass of CO₂ = 12 (C) + 2 × 16 (O) = 44 g/mol

• Mass of CO₂ = Moles × Molar mass
  = 0.5 mol × 44 g/mol = 22 g

Explanation:

Balanced equation:
2KClO₃ → 2KCl + 3O₂
From the equation:
2 moles of KClO₃ → 3 moles of O₂
This means:
1 mole of KClO₃ → 1.5 moles of O₂
So, for 10 moles of KClO₃:
10 × 1.5 = 15 moles of O₂

Molar mass of Al= 27g/mol,  Molar mass of Al2O3= 102g/mol

Moles of Al= 18.5/27= 0.6mol

Moles of Al2O3= 2/4 (0.6)= 0.34mol

Mass of Al2O3= (0.34)(102)= 34.9g

ZnCO₃→ZnO+CO₂
From the balanced equation, 1 mole of ZnCO₃ produces 1 mole of CO₂. Therefore, 10 moles of ZnCO₃ will produce 10 moles of CO₂.

The molar mass of CO₂ is: C=12g/mol,
O₂=16×2=32g/mol
⇒
Molar mass of CO₂=12+32=44g/mol
C=12g/mol,O₂=16×2=32g/mol ⇒Molar mass of CO₂=12+32=44g/mol
For 10 moles of CO₂:

Mass of CO₂
=10moles×44g/mol=440g

Explanation:

Reaction needs 2 mol H₂ for 1 mol O₂. For 1 mol O₂, required H₂ = 2 mol. We have 4 mol H₂ (extra). O₂ finishes first, so O₂ is limiting.

The balanced reaction is as follows:
                                    $2A+4B\to 3C+4D$

No. of moles  given: $5,6$       

Stoichiometry coefficients given: $2,4$       

Ratio of moles given and stoichiometry coefficients:    

KPK Book Example

If the mass of $\mathrm{HCl}$ is doubled from 15.5 g to 31.0 g, then the moles of $\mathrm{HCl}$ will also double:

Moles=31/36.5
moles=0.850 mol

If 4.4 moles of ZnCO3​ are decomposed, then 4.4 moles of CO2​ will be produced, as the mole ratio is 1:1. To find the mass:

$$4.4\text{moles}\times 44\text{g/mol}=193.6\text{g}$$

From the equation:
2 moles of KClO₃ produce:
→ 2 moles of KCl
→ 3 moles of O₂

This gives us the mole ratio: KCl:O2 =2 :3

The reaction requires equal moles of iron and sulphur (1:1 ratio).
Given 56 g of iron (1 mole) reacts with only $\text{g}$ of sulphur (1 mole),
leaving 64−32=32 g of sulphur in excess.

Reference: KPK Book Example

The reaction requires equal moles of iron and sulphur to form iron sulphide.

Since the reaction will limit to the smaller stoichiometric ratio (2 moles of iron), it will react with 2 moles of sulphur (64 g). Therefore, the mass of iron sulphide produced is:  2 × 88 =176 g

The limiting reagent in a chemical reaction is the reactant that will be consumed completely. Once there is no more of that reactant, the reaction cannot proceed. Therefor it limits the reaction from continuing. The excess reagent is the reactant that could keep reacting if the other had not been consumed.

The balanced chemical equation shows that 2 moles of KClO₃ produce 3 moles of O₂. Therefore, 5 moles of KClO₃ will produce:
moles of O2=(3/2)×5=7.5 moles

The limiting reagent controls the amount of product formed in a chemical reaction.

The limiting reagent (or limiting reactant) is the substance that is completely consumed first in a reaction, thereby determining the maximum amount of product that can be formed. Once the limiting reagent is used up, the reaction stops, and no more product can be produced, even if other reactants are still available in excess.

Explanation:

In this process, glucose (solute) is dissolved in water (solvent) to form a solution.
The limiting reagent is the substance that is completely used up first — here, glucose — because water is present in large excess and does not limit the formation of the solution.

Convert Gram into moles and 
from balance chemical equation we can see that

now 120 g of CO2 produced a smaller number of moles of H2CO3 so it is limiting reagent

Write the balanced decomposition reaction:
ZnCO₃ → ZnO + CO₂

From the equation,
1 mol ZnCO₃ produces 1 mol ZnO.

4 mol ZnCO₃ × = 4 mol ZnO

The chemical equation of water formation is 2H2 + O2 → 2H2O. Though we have 8g of hydrogen, here oxygen is the limiting reagent. So only 4g of hydrogen can be used to produce water i.e. 36g of water. That is 2 moles.

Though the other substances are excess in amount than the required, each and every reactant needs to be in a fixed ratio to attain the desired product. So, thereby, the reactant that limits the quantity of the product formed is called limiting reagent and this reactant gets consumed first completely.

Limiting reactant in any reaction is the one that is consumed earlier