Answer:
The balanced chemical equation for the combustion of propane is:
C3H8 + 5 O2 → 3 CO2 + 4 H2O
From the equation, we see that for every 1 mole of propane burned, 3 moles of carbon dioxide are produced. Therefore, we need to first convert the given mass of propane to moles.
The molar mass of propane is:
3 x 12.01 (atomic mass of carbon) + 8 x 1.01 (atomic mass of hydrogen) = 44.1 g/mol
200.0 g of propane is therefore equivalent to:
200.0 g / 44.1 g/mol = 4.53 moles of propane
Since 3 moles of CO2 are produced for every mole of propane burned, the number of moles of CO2 produced is:
4.53 moles of propane x 3 moles of CO2 / 1 mole of propane = 13.6 moles of CO2
To convert moles of CO2 to liters of CO2 at STP (standard temperature and pressure), we can use the ideal gas law:
PV = nRT
Where:
P = pressure = 103 kPa
V = volume in liters (what we're solving for)
n = number of moles of gas = 13.6 mol
R = gas constant = 0.08206 L·atm/mol·K
T = temperature in Kelvin = 23.0°C + 273.15 = 296.15 K
Solving for V, we get:
V = nRT/P = (13.6 mol) x (0.08206 L·atm/mol·K) x (296.15 K) / (103 kPa)
V ≈ 381 liters
Therefore, 200.0 grams of propane would produce approximately 381 liters of carbon dioxide at 23.0°C and 103 kPa
Step-by-step explanation:
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