When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s) + CO₂(g) What is the mass of calcium carbonate needed to produce 35.0 L of carbon dioxide at STP? Express your answer with the appropriate units. View Available Hint(s) 0 mass of CaCO3 = 370.8 Part B μᾶ atm Submit Previous Answers Request Answer Incorrect; Try Again; 4 attempts remaining 2 Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10 (9) + 1302 (g)→8CO2 (g) + 10H₂O(1) At 1.00 atm and 23 °C, what is the volume of carbon dioxide formed by the combustion of 2.20 g of butane?
The Correct Answer and Explanation is :
To solve the problems you have outlined, let’s break them down one by one.
Part A: Mass of Calcium Carbonate needed to produce 35.0 L of Carbon Dioxide at STP
The decomposition reaction of calcium carbonate (CaCO₃) is:
[ \text{CaCO₃ (s)} \rightarrow \text{CaO (s)} + \text{CO₂ (g)} ]
To calculate the mass of calcium carbonate needed to produce 35.0 L of CO₂ at STP, we will follow these steps:
- Use the ideal gas law to find moles of CO₂ produced:
[
PV = nRT
]
At STP (Standard Temperature and Pressure), the molar volume of a gas is 22.4 L. Therefore, the moles of CO₂ can be calculated as:
[
n_{\text{CO₂}} = \frac{V_{\text{CO₂}}}{22.4 \, \text{L/mol}} = \frac{35.0 \, \text{L}}{22.4 \, \text{L/mol}} = 1.56 \, \text{mol CO₂}
] - Use stoichiometry to determine moles of CaCO₃:
The balanced reaction shows that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore:
[
n_{\text{CaCO₃}} = 1.56 \, \text{mol CaCO₃}
] - Calculate the mass of CaCO₃:
The molar mass of CaCO₃ is approximately 100.1 g/mol. So, the mass of CaCO₃ required is:
[
\text{Mass of CaCO₃} = n_{\text{CaCO₃}} \times \text{Molar mass of CaCO₃}
]
[
\text{Mass of CaCO₃} = 1.56 \, \text{mol} \times 100.1 \, \text{g/mol} = 156.2 \, \text{g}
]
Thus, 156.2 g of calcium carbonate is needed to produce 35.0 L of CO₂ at STP.
Part B: Volume of CO₂ produced by combustion of 2.20 g of Butane
The balanced equation for the combustion of butane (C₄H₁₀) is:
[ 2 \, \text{C₄H₁₀ (g)} + 13 \, \text{O₂ (g)} \rightarrow 8 \, \text{CO₂ (g)} + 10 \, \text{H₂O (l)} ]
- Calculate the moles of butane (C₄H₁₀):
The molar mass of butane is approximately 58.12 g/mol. Using the given mass of butane:
[
n_{\text{C₄H₁₀}} = \frac{\text{Mass of C₄H₁₀}}{\text{Molar mass of C₄H₁₀}} = \frac{2.20 \, \text{g}}{58.12 \, \text{g/mol}} = 0.0379 \, \text{mol C₄H₁₀}
] - Use stoichiometry to find the moles of CO₂:
From the balanced equation, we know that 2 moles of C₄H₁₀ produce 8 moles of CO₂. Therefore:
[
n_{\text{CO₂}} = \frac{8}{2} \times n_{\text{C₄H₁₀}} = 4 \times 0.0379 \, \text{mol} = 0.1516 \, \text{mol CO₂}
] - Use the ideal gas law to calculate the volume of CO₂ at the given conditions:
At 1.00 atm and 23°C (which is 296 K), the molar volume of an ideal gas can be calculated using the ideal gas law:
[
V = nRT / P
]
Where:
- (n = 0.1516 \, \text{mol CO₂})
- (R = 0.0821 \, \text{L·atm/(mol·K)})
- (T = 296 \, \text{K})
- (P = 1.00 \, \text{atm}) Substituting the values:
[
V = \frac{(0.1516 \, \text{mol}) \times (0.0821 \, \text{L·atm/mol·K}) \times (296 \, \text{K})}{1.00 \, \text{atm}} = 3.69 \, \text{L}
]
Thus, 3.69 L of CO₂ is produced by the combustion of 2.20 g of butane at 1.00 atm and 23°C.
Visual Representation
Now, I will generate an image representing the steps involved in these calculations.
Here is the visual representation of the calculations for both problems. The diagram outlines the chemical reactions, stoichiometry, and conversion factors for each, along with the steps to calculate the mass of calcium carbonate and the volume of carbon dioxide produced. Let me know if you need further clarification on any part of the solution!
