Stoichiometry And Stoichiometric Calculations

To determine how many grams of are produced from the complete combustion of 10 g of (ethane), we follow these steps:

Step 1: Determine the molar mass of and .

• Molar mass of : (2 x 12.01 g/mol for C) + (6 x 1.008 g/mol for H) = 30.07 g/mol
• Molar mass of : (1 x 12.01 g/mol for C) + (2 x 16.00 g/mol for O) = 44.01 g/mol

Step 2: Calculate the moles of in 10 g.

Moles of = ≈ 0.332 mol

Step 3: Use the stoichiometry of the reaction to find moles of produced.

The balanced reaction is:

The molar ratio of to is 2:4 or 1:2. Therefore, 0.332 mol of will produce 0.664 mol of .

Step 4: Calculate the grams of produced.

Mass of = 0.664 mol x 44.01 g/mol = 29.23 g

Therefore, the mass of produced is approximately '29.3 g'.

To find the volume of CO₂ gas produced at STP (Standard Temperature and Pressure) when 10 g of CaCO₃ reacts with excess HCl, follow these steps:

1. Start with the balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl):

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

2. Calculate the molar mass of CaCO₃:

Molar mass of CaCO₃ = (40.08 g/mol for Ca) + (12.01 g/mol for C) + (3 × 16.00 g/mol for O)
Molar mass of CaCO₃ = 100.09 g/mol

3. Determine the number of moles of CaCO₃ in 10 g:

Number of moles = 10 g / 100.09 g/mol = 0.0999 mol

4. According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.0999 moles of CaCO₃ will produce 0.0999 moles of CO₂.
5. At STP, 1 mole of gas occupies 22.4 L. Calculate the volume of CO₂ produced:

Volume of CO₂ = 0.0999 mol × 22.4 L/mol = 2.2368 L

Rounding to appropriate significant figures, the volume is approximately 2.24 L.

Therefore, the volume of CO₂ gas produced at STP is 2.24 L, which matches the given option.

To determine the moles of oxygen required for the complete combustion of 1 mole of propane (C H ), we must first write the balanced chemical equation for the combustion process. The combustion of propane is represented by:

C H + 5O → 3CO + 4H O

This equation indicates that 1 mole of propane (C H ) reacts with 5 moles of oxygen (O ) to produce 3 moles of carbon dioxide (CO ) and 4 moles of water (H O).

Therefore, 5 moles of oxygen are required to completely combust 1 mole of propane.

To determine the number of moles of CO₂ produced when 2 moles of C₂H₆ are completely burnt, we must first consider the balanced chemical equation for the combustion of ethane (C₂H₆).

The balanced equation for the complete combustion of ethane is:

2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

From the equation, we can see that 2 moles of C₂H₆ produce 4 moles of CO₂ when completely burnt in oxygen.

Given that we start with 2 moles of C₂H₆, using stoichiometry as per the balanced equation, we calculate the moles of CO₂ produced:

• According to the balanced equation, the molar ratio of C₂H₆ to CO₂ is 2:4, or 1:2.
• Thus, 2 moles of C₂H₆ produce moles of CO₂.

Therefore, the number of moles of CO₂ produced is 4.