Consider the burning of LPG (propane C3H8)which produces carbon dioxide (CO2) and water vapor (H2O). The balanced chemical equation involving the reaction is C3H8+5O2= 3CO2+4H2O
Suppose 1.5 moles of C3H8 will be used, assume the following:
a. How many moles of O2 will be needed?
b. How many moles of H2O and CO2 will be formed?
c. How many grams will 1.5 moles C3H8 makes?
d. What volume (at STP)of O2 and CO2 are involved?
e. How many grams of water vapor will be formed?
C3H8 + 5O2 => 3CO2 + 4H2O
a. Use the coefficients in the balanced equation to convert anything to anything.
1.5 mols C3H8 x (5 mols O2/1 mol C3H8) = ?
b. same process.
c. grams = mols x molar mass but you didn't say grams of what.
d. Remember 1 mol of anything occupies 22.4 L at STP.
e. g H2O = mols H2O x molar mass H2O.
1.35g
To solve this problem, we will use the balanced chemical equation provided: C3H8 + 5O2 = 3CO2 + 4H2O
a. How many moles of O2 will be needed?
According to the balanced equation, for every 1 mole of C3H8, we need 5 moles of O2. Since we have 1.5 moles of C3H8, we multiply it by the ratio: (1.5 moles C3H8) * (5 moles O2 / 1 mole C3H8) = 7.5 moles of O2.
b. How many moles of H2O and CO2 will be formed?
Again, referring to the balanced equation, we know that 1 mole of C3H8 produces 3 moles of CO2 and 4 moles of H2O. Since we have 1.5 moles of C3H8, we calculate: (1.5 moles C3H8) * (3 moles CO2 / 1 mole C3H8) = 4.5 moles CO2, and (1.5 moles C3H8) * (4 moles H2O / 1 mole C3H8) = 6 moles H2O.
c. How many grams will 1.5 moles C3H8 make?
To calculate this, we need the molar mass of C3H8 (propane). The molar mass of C3H8 is: (3 * molar mass of C) + (8 * molar mass of H) = (3 * 12.01 g/mol) + (8 * 1.01 g/mol) = 44.11 g/mol.
Therefore, 1.5 moles of C3H8 will have a mass of: (1.5 moles C3H8) * (44.11 g/mol) = 66.16 grams.
d. What volume (at STP) of O2 and CO2 are involved?
To find the volume, we need to use the Ideal Gas Law: PV = nRT.
At STP (Standard Temperature and Pressure), T = 273.15 K, P = 1 atm, and R = 0.0821 L·atm/(mol·K).
For O2:
From part (a), we know that we have 7.5 moles of O2. Plugging this into the Ideal Gas Law, we can solve for V:
V = (nRT) / P = (7.5 moles) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm) = 172.56 L.
For CO2:
From part (b), we know that we produce 4.5 moles of CO2. Similar to O2, we use the Ideal Gas Law:
V = (nRT) / P = (4.5 moles) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm) = 91.11 L.
e. How many grams of water vapor will be formed?
To find the mass of water vapor (H2O) formed, we need to use the molar mass of H2O, which is 18.02 g/mol. From part (b), we know that we produce 6 moles of H2O. The mass of water vapor formed is: (6 moles H2O) * (18.02 g/mol) = 108.12 grams.