How do I calculate this problem? How much water vapor is produced for each amount of propane burned. Mass of c3h8 reacted 44, mass of O2 reacted 160, mass of CO2 produces 132?
To calculate the amount of water vapor produced when burning propane, we can use the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 → 3CO2 + 4H2O
We are given the masses of propane (C3H8) reacted, oxygen (O2) reacted, and carbon dioxide (CO2) produced. However, we need to find the mass of water vapor (H2O) produced.
First, let's find the molar mass of each compound. The molar mass of C3H8 is 44 g/mol, O2 is 32 g/mol (16 g/mol for each oxygen atom), CO2 is 44 g/mol (12 g/mol for carbon + 32 g/mol for oxygen), and H2O is 18 g/mol (2 g/mol for hydrogen + 16 g/mol for oxygen).
Now, let's calculate the number of moles for each compound using the given masses:
Number of moles of C3H8 = Mass of C3H8 / Molar mass of C3H8
Number of moles of O2 = Mass of O2 / Molar mass of O2
Number of moles of CO2 = Mass of CO2 / Molar mass of CO2
Using the given values, we have:
Number of moles of C3H8 = 44 g / 44 g/mol = 1 mole
Number of moles of O2 = 160 g / 32 g/mol = 5 moles
Number of moles of CO2 = 132 g / 44 g/mol = 3 moles
According to the balanced equation, for every 1 mole of C3H8 reacted, we get 4 moles of H2O produced. Therefore, the number of moles of H2O produced can be calculated using the stoichiometry:
Number of moles of H2O = (Number of moles of C3H8) × (4 moles of H2O / 1 mole of C3H8)
Plugging in the values, we get:
Number of moles of H2O = 1 mole × (4 moles of H2O / 1 mole of C3H8) = 4 moles
Finally, let's calculate the mass of H2O produced using the molar mass of H2O:
Mass of H2O = Number of moles of H2O × Molar mass of H2O
Mass of H2O = 4 moles × 18 g/mol = 72 grams
Therefore, for the given masses of reactants and products, 72 grams of water vapor is produced when burning the indicated amounts of propane.
To calculate the amount of water vapor produced for each amount of propane burned, we need to use balanced chemical equations and stoichiometry.
The balanced chemical equation for the reaction of propane (C3H8) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the equation, we can see that for every 3 moles of CO2 produced, 4 moles of H2O are also produced.
First, let's calculate the number of moles of propane burned:
44 g of C3H8 reacted / molar mass of C3H8 = moles of C3H8 reacted
The molar mass of C3H8 (propane) is:
(3 x atomic mass of carbon) + (8 x atomic mass of hydrogen)
= (3 x 12.01 g/mol) + (8 x 1.01 g/mol)
= 36.03 g/mol + 8.08 g/mol
= 44.11 g/mol
So, moles of C3H8 reacted = 44 g / 44.11 g/mol = 1 mol
Next, let's calculate the moles of water vapor produced:
For every 3 moles of CO2 produced, 4 moles of H2O are produced.
From the given information, we know that 132 g of CO2 are produced, so we can calculate the moles of CO2 produced:
132 g of CO2 produced / molar mass of CO2 = moles of CO2 produced
The molar mass of CO2 is:
(1 x atomic mass of carbon) + (2 x atomic mass of oxygen)
= 12.01 g/mol + (2 x 16.00 g/mol)
= 12.01 g/mol + 32.00 g/mol
= 44.01 g/mol
So, moles of CO2 produced = 132 g / 44.01 g/mol = 3 mol
Now, we can calculate the moles of water vapor produced:
moles of H2O produced = (moles of CO2 produced) x (moles of H2O produced per mole of CO2)
moles of H2O produced = 3 mol x (4 mol H2O / 3 mol CO2) = 4 mol
Finally, we can calculate the mass of water vapor produced:
mass of H2O produced = (moles of H2O produced) x (molar mass of H2O)
The molar mass of H2O (water) is:
(2 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)
= (2 x 1.01 g/mol) + 16.00 g/mol
= 2.02 g/mol + 16.00 g/mol
= 18.02 g/mol
So, mass of H2O produced = 4 mol x 18.02 g/mol = 72.08 g
Therefore, for the given amounts of propane burned (44 g) and carbon dioxide produced (132 g), the mass of water vapor produced is 72.08 g.