Stoichiometry And Stoichiometric Calculations

The correct answer is 18
44g of CO₂ contains 12g of carbon
0.20g of CO₂ will contain x g
Therefore ,

Percentage =
= 18 %

The combustion reaction can be represented as:
Given that 9.5 moles of are required and 3 moles of water are produced, we can set up the following equations:
-
-
Thus, the molecular formula of the hydrocarbon is .
This corresponds to an alkyne or aromatic compound.

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Given Mass of glucose = 900 gram

Molar mass of glucose = 180 gm/mole

So, moles of glucose:

moles of glucose ⇒ 900/180 = 5

1 mole of glucose requires 6 moles of oxygen gas.

So, 5 moles of glucose will require 30 moles of oxygen gas.

Moles of oxygen gas = 30

Molar mass of oxygen gas = 32 g/mole

So, mass of oxygen gas required:

mass of oxygen gas required ⇒ 30 × 32 = 960 gram

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93 g of aniline produces 330 g of 2,4,6-tribromoaniline.
Theoretical yield for 9.3 g of aniline:

Percentage yield:

The reaction for the nitration of nitrobenzene to m-dinitrobenzene is: Nitrobenzene (C H NO ) has a molar mass (MW) of: m-dinitrobenzene (C H N O ) has a molar mass (MW) of: From the stoichiometry of the reaction, 1 mole of nitrobenzene produces 1 mole of m-dinitrobenzene.
Moles of m-dinitrobenzene produced = Since the mole ratio of nitrobenzene to m-dinitrobenzene is 1:1, the moles of nitrobenzene reacted are also 0.025 mol.
Mass of nitrobenzene reacted (X) = moles of nitrobenzene × molar mass of nitrobenzene The nearest integer to 3.075 is 3. Therefore, X = 3 g.

To determine the molarity of the sodium iodide (NaI) solution, we'll follow these steps:

1. Write the Reaction Equation:

The reaction between sodium iodide (NaI) and silver nitrate (AgNO₃) produces silver iodide (AgI) and sodium nitrate (NaNO₃):

NaI + AgNO₃ → AgI + NaNO₃

2. Calculate Moles of Silver Iodide:

The molar mass of AgI = Ag + I = 108 + 127 = 235 g/mol. Given silver iodide mass is 4.74 g:

Moles of AgI = 4.74 g / 235 g/mol = 0.02017 mol

3. Relate Moles to Sodium Iodide:

From the balanced equation, moles of NaI = moles of AgI = 0.02017 mol

4. Calculate Molarity of NaI:

Volume of NaI solution = 20 mL = 0.020 L.

Molarity (M) = Moles of solute / Volume of solution in liters = 0.02017 mol / 0.020 L = 1.0085 M

Conclusion:

The nearest integer molarity of the sodium iodide solution is 1 M, which falls within the range of 1 to 1. Thus, the molarity is confirmed as 1 M.

Given that the molar mass of sulphur (S) is 32 g/mol and that of barium sulphate (BaSO ) is 233 g/mol, the mass of sulphur in the compound can be calculated from the mass of barium sulphate produced:
The moles of barium sulphate formed:
The moles of sulphur in the compound are equal to the moles of BaSO because of the 1:1 stoichiometry of BaSO and sulphur.
The mass of sulphur is:
The percentage of sulphur in the compound is: Thus, the correct percentage of sulphur is

Organic Compound → CO₂

Applying POAC on ‘C’

Option (mole) of ‘C’ in compound =

So mass of ‘C’ in compound =

So, % of ‘C’ in compound =

= 79.63

To determine the mass of needed to achieve 12 ppm of iron in 150 kg of wheat, follow these steps:

Define ppm: 1 ppm is 1 mg of solute per 1 kg of solution. Therefore, for 12 ppm of iron in 150 kg of wheat, the mass of iron needed is:

To find the mass of , start by calculating its molar mass:

Calculate the moles of in 1.8 g:

Convert moles of to moles of since each unit contains one Fe atom:

Find the mass of :

Round to the nearest integer: 9 g

Verify the solution is within the range 9–9. It is correct.

Therefore, the mass of required is 9 g.

To find the mass of aluminium oxide (Al₂O₃) produced, we start with the balanced chemical equation for the reaction:

4Al + 3O₂ → 2Al₂O₃

1. **Molar Mass Calculation:**
- Al (Aluminium) = 27 g/mol
- O₂ (Oxygen) = 32 g/mol
- Al₂O₃ (Aluminium oxide) = (2×27 + 3×16) g/mol = 102 g/mol

2. **Determine Limiting Reagent:**

The moles of Al from 81.0 g:

n(Al) = 81.0 g / 27 g/mol = 3.0 mol

The moles of O₂ from 128.0 g:

n(O₂) = 128.0 g / 32 g/mol = 4.0 mol

According to the equation, 4 moles of Al react with 3 moles of O₂. Calculate the required oxygen for 3.0 mol of Al:

Required O₂ = (3.0 mol Al) × (3/4) = 2.25 mol O₂

Since 2.25 mol of O₂ is less than the available 4.0 mol, Al is the limiting reagent.

3. **Calculate Mass of Al₂O₃ Produced:**

From 3.0 mol of Al (1.5 mol of Al₂O₃):

n(Al₂O₃) = 3.0 mol × (2/4) = 1.5 mol

Mass of Al₂O₃ = 1.5 mol × 102 g/mol = 153.0 g

The mass of aluminium oxide produced is 153 g.

This value falls within the expected range of 153 to 153 g.

To determine the amount of calcium oxide (CaO) produced from heating 150 kg of limestone (CaCO₃), knowing the limestone is 75% pure, we follow these steps:

1. Calculate the mass of pure CaCO₃: The limestone is 75% pure, so the mass of pure CaCO₃ is 75% of 150 kg.
Mass of pure CaCO₃ = 0.75 × 150 kg = 112.5 kg.

2. Use the chemical reaction:
CaCO₃ (s) → CaO (s) + CO₂ (g)

3. Calculate the molar masses:
Ca = 40, C = 12, O = 16 (given).
Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol.
Molar mass of CaO = 40 + 16 = 56 g/mol.

4. Convert mass of CaCO₃ to moles:
Moles of CaCO₃ = = 1125 moles.

5. Moles of CaCO₃ to moles of CaO: The reaction shows a 1:1 molar ratio, so moles of CaO = 1125 moles.

6. Calculate the mass of CaO:
Mass of CaO = moles of CaO × molar mass of CaO = 1125 moles × 56 g/mol = 63000 g = 63 kg.

7. Verify the range: The computed mass of CaO is 63 kg, which fits perfectly within the given range of 63-63 kg.

Therefore, the amount of calcium oxide produced is 63 kg.

We are asked to calculate the mass of calcium chloride ( ) formed from the reaction between 250 mL of 0.76 M hydrochloric acid ( ) and 1000 g of calcium carbonate ( ). The balanced chemical equation is provided.

Concept Used:

This is a stoichiometry problem that involves a limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction and therefore determines the maximum amount of product that can be formed.

1. Calculate the moles of each reactant from the given quantities.
2. Use the stoichiometry of the balanced equation to determine which reactant is the limiting one.
3. Calculate the moles of the product formed based on the moles of the limiting reactant.
4. Convert the moles of the product to mass using its molar mass.

The formula for moles from molarity and volume is: . The formula for moles from mass is: .

Step-by-Step Solution:

Step 1: Calculate the molar masses of the reactants and the product.

Given molar masses: Ca = 40, C = 12, O = 16, Cl = 35.5 g/mol.

• Molar mass of .
• Molar mass of .

Step 2: Calculate the number of moles of each reactant.

For HCl:

For :

Step 3: Identify the limiting reactant.

The balanced chemical equation is: .

The stoichiometry shows that 1 mole of reacts with 2 moles of .

Let's determine how many moles of are required to react completely with the available :

We have 10 moles of available, which is much more than the 0.095 moles needed. Therefore, is in excess, and HCl is the limiting reactant.

Step 4: Calculate the moles of formed.

The amount of product formed is determined by the limiting reactant (HCl). From the balanced equation, 2 moles of produce 1 mole of .

Step 5: Calculate the mass of formed.

Using the molar mass of calculated in Step 1:

The mass of formed will be 10.545 g.

To determine the mass of magnesium (Mg) required to produce 220 mL of hydrogen gas (H₂) at Standard Temperature and Pressure (STP) when reacted with excess dilute hydrochloric acid (HCl), we start by considering the balanced chemical reaction:

1. The balanced equation shows that 1 mole of magnesium produces 1 mole of hydrogen gas.
2. At STP, 1 mole of any ideal gas occupies 22.4 L (or 22,400 mL).
3. Given that 220 mL of hydrogen is produced, we calculate the moles of hydrogen gas:

1. Since 1 mole of Mg produces 1 mole of H₂, the moles of magnesium required is also 0.00982 mol.
2. The molar mass of magnesium (Mg) is given as 24 g/mol. Thus, the mass of magnesium required is:

1. Converting grams to milligrams (1 g = 1000 mg):

Thus, the mass of magnesium required is approximately 236 mg.

Conclusion: The correct answer is 236 mg.

The percentage of sulfur in an organic compound can be estimated using the Carius method. In this method, the sulfur present in the organic compound is converted to barium sulfate ( ).

1. Molar Mass Calculations:
The molar mass of is calculated as:
g/mol.
The molar mass of S is 32 g/mol.

2. Stoichiometric Relationship:
1 mole of contains 1 mole of S.
So, 233 g of contains 32 g of S.

3. Experimental Data:
Given that 160 mg of organic compound gives 466 mg of .
The amount of sulfur in 466 mg of is:

The mass of sulfur in the organic compound is 64 mg.

4. Percentage Calculation:
The percentage of sulfur in the organic compound is:

Final Answer:
The percentage of sulfur in the given compound is 40%.

The final answer is .

To determine the amount of bromine required to convert 2 g of phenol into 2,4,6-tribromophenol, we need to consider the balanced chemical reaction and the stoichiometry involved. Let's go through the steps:

1. The reaction between phenol and bromine ( ) to form 2,4,6-tribromophenol is as follows:
   
2. Determine the molar masses:
   • Molar mass of phenol ( ):
     Carbon (C): 12 × 6 = 72 g/mol
     Hydrogen (H): 1 × 6 = 6 g/mol
     Oxygen (O): 16 × 1 = 16 g/mol
     Total = 72 + 6 + 16 = 94 g/mol
   • Molar mass of bromine ( ):
     Bromine (Br): 80 × 2 = 160 g/mol
3. Determine the moles of phenol:
   
4. According to the balanced equation, 1 mole of phenol reacts with 3 moles of bromine. Therefore, the moles of bromine required will be:
   
5. Calculate the mass of bromine needed:
   

Therefore, the amount of bromine required is approximately 10.22 g, corresponding to the correct option: 10.22 g. This calculation confirms that option is correct.

Step 1: Calculating moles of aniline.
Molar mass of aniline = 93 g mol⁻¹

Step 2: Stoichiometry of the reaction.
1 mole of aniline produces 1 mole of acetanilide.
Molar mass of acetanilide = 135 g mol⁻¹.
Step 3: Theoretical yield of acetanilide.

Step 4: Percentage yield calculation.