Stoichiometry And Stoichiometric Calculations

Given: - 12 moles of NH3 - , , First, calculate the molar mass of HNO3 (Nitric Acid): Next, find the moles of nitric acid formed, which will be equal to the moles of NH3 oxidized, assuming 1 mole of NH3 produces 1 mole of HNO3: Now, calculate the mass of HNO3 produced:

Step 1: Determine the Empirical Formula:
Let the mass of hydrocarbon be 100g.

• Carbon: 75g. Moles of C = .
• Hydrogen: . Moles of H = .

Mole Ratio C : H = . Empirical Formula is (Methane). Molar Mass of g/mol.
Step 2: Calculate Moles of Hydrocarbon Used:
Given Mass = 32g. Moles of moles.
Step 3: Combustion Stoichiometry:
Reaction: . From stoichiometry, 1 mole of produces 1 mole of . Therefore, 2 moles of will produce 2 moles of .
Step 4: Calculate Mass of :
Molar Mass of g/mol. Mass of produced = g.
Step 5: Convert to kg:
Question asks for weight in kg. .

Step 1: Calculate the initial moles of the reactants.
The number of moles is (in Liters). Moles of : . Moles of (which provides ): .

Step 2: Write the balanced redox reaction equation.
is reduced by (iodide) to (which precipitates as ), and is oxidized to .

Step 3: Identify the limiting reagent.
From the stoichiometry of the reaction, moles of require moles of . The required mole ratio is . Available mole ratio: . Since the available ratio (20) is much greater than the required ratio (2), is the limiting reagent and will be completely consumed.

Step 4: Calculate the moles of products formed.
Based on the limiting reagent ( ): From the balanced equation: Moles of : . Moles of :

Step 5: Conclude the final answer.
The number of moles of and formed are moles and moles respectively.

We need to calculate the oxidation state of Nitrogen (N) in each compound. Let the oxidation state of N be .
We use the standard rules for assigning oxidation states: Oxygen is usually -2, Hydrogen is usually +1, and Chlorine (as a halide ion) is -1. The sum of oxidation states in a neutral compound is 0.
(A) NH OH (Hydroxylamine): The compound can be seen as NH and OH. . . .
(B) NH Cl (Ammonium chloride): This is an ionic compound composed of the ammonium ion (NH ) and the chloride ion (Cl ). We calculate the oxidation state of N in NH . . .
(C) N H (Hydrazine): . . .
(D) HNO (Nitrous acid): . . .
Comparing the oxidation states of Nitrogen in each compound: (A) -1 (B) -3 (C) -2 (D) +3
The lowest (most negative) oxidation state is -3, which is found in NH Cl.

Step 1: Understanding the Concept:
This is a problem based on stoichiometry and thermal decomposition. When the mixture of sodium carbonate (Na₂CO₃) and sodium bicarbonate (NaHCO₃) is heated, only sodium bicarbonate decomposes. Sodium carbonate is thermally stable at 673K and does not decompose. The loss in mass is due to the gaseous products (CO₂ and H₂O) escaping.
Step 2: Key Formula or Approach:
The balanced chemical equation for the decomposition of NaHCO₃ is:
We will use this equation to relate the mass of NaHCO₃ decomposed to the mass of the gaseous products lost.
Step 3: Detailed Explanation:
First, calculate the molar masses of the relevant compounds:
- Molar mass of NaHCO₃ = 23 + 1 + 12 + 3(16) = 84 g/mol.
- Molar mass of H₂O = 2(1) + 16 = 18 g/mol.
- Molar mass of CO₂ = 12 + 2(16) = 44 g/mol.
From the balanced equation, the decomposition of 2 moles of NaHCO₃ results in the loss of 1 mole of H₂O and 1 mole of CO₂.
- Mass of 2 moles of NaHCO₃ = g.
- Total mass of gaseous products lost = Mass of H₂O + Mass of CO₂ = g.
This means that 168 g of NaHCO₃ produces a mass loss of 62 g.
We are given that the actual loss in mass is 0.62 g. We can set up a proportion to find the mass of NaHCO₃ in the original mixture.
Let be the mass of NaHCO₃ in the mixture.
So, the mass of NaHCO₃ in the mixture is 1.68 g.
The total mass of the mixture is 4.0 g.
Mass of Na₂CO₃ = Total mass - Mass of NaHCO₃ = .
Finally, calculate the percentage of sodium carbonate (Na₂CO₃) in the mixture:
Step 4: Final Answer:
The percentage of sodium carbonate in the mixture is 58%. Therefore, option (B) is correct.

• Total moles KI = mol
• Moles Cu = mol
• Each Cu reacts with 2 KI → 0.01 mol KI consumed.
• Remaining KI = mol in L → M.

1. The reaction is CaCO CaO + CO (decomposition by burning).
2. Purity 90%, so effective mass = 0.9 \times 10 = 9 g.
3. Molar mass CaCO = 100 g/mol, moles = 9 / 100 = 0.09 mol.
4. 1 mol CaCO produces 1 mol CO , so 0.09 mol CO .
5. At STP, volume = moles \times 22.7 L/mol = 0.09 \times 22.7 \approx 2.043$ L \approx 2.044 L.
6. Therefore, the correct option is (3) 2.044 L.