Question

Определите объём углекислого газа который образуется при сгорании 4,8 г угля

Answer 1

C + O2 = CO2
n(C) = 4,8/12 = 0,4 моль
V(CO2) = 0,4*22,4 = 8,96 л

Answer 2

Calculating the Volume of Carbon Dioxide Produced from the Combustion of Coal

When 4.8 grams of coal is burned, we can calculate the volume of carbon dioxide (CO2) produced using the following steps:

1. Determine the Chemical Equation for the Combustion of Coal: - The chemical equation for the combustion of coal can be represented as: ``` C + O2 -> CO2 ``` - This equation shows that one mole of carbon (C) reacts with one mole of oxygen (O2) to produce one mole of carbon dioxide (CO2).

2. Calculate the Moles of Carbon in 4.8 grams of Coal: - The molar mass of carbon (C) is approximately 12.01 g/mol. - Using the formula: moles = mass / molar mass, we can calculate the moles of carbon in 4.8 grams of coal. - This gives us: moles of carbon = 4.8 g / 12.01 g/mol.

3. Determine the Volume of Carbon Dioxide Produced: - Since 1 mole of carbon produces 1 mole of carbon dioxide, the moles of carbon dioxide produced will be the same as the moles of carbon. - We can then use the ideal gas law, PV = nRT, to calculate the volume of carbon dioxide produced, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

4. Substitute Values and Calculate: - We can use the ideal gas constant, R, which is approximately 0.0821 L·atm/mol·K, and assume standard temperature and pressure (STP) for the calculation (0°C or 273.15 K and 1 atm). - Using the ideal gas law equation, V = nRT/P, we can calculate the volume of carbon dioxide produced.

Calculation

- Moles of carbon = 4.8 g / 12.01 g/mol = 0.399 moles - Using the ideal gas law at STP: V = nRT/P - V = (0.399 moles) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)

Result

- The volume of carbon dioxide produced from the combustion of 4.8 grams of coal at STP is approximately 8.87 liters.