Let's see.
C3H8 + 5O2 -> 3CO2 + 4H2O
For every one molecule of propane burned there is four molecules of water produced.
Or, this is the actuality.
1 molecule propane (1 mole C3H8/6.022 X 1023)(4 mole H2O/ 1mole C3O8)
= 6.64 X 10 -24 molecules water
It depends on the mass of propane burned (this should have been obvious).
Assuming complete combustion, 44 grams of propane burns to produce 72 grams of water (plus some carbon dioxide). The extra mass comes from atmospheric oxygen.
1
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
Assuming that the water is produced by complete combustion of propane, the balanced equation is: C3H8 + 5 O2 = 3 CO2 + 4 H2O. The gram molecular mass of propane is three times the gram atomic mass of carbon plus eight times the gram atomic mass of hydrogen = 3(12.011) + 8(1.008) = 44.097, and the gram molecular mass of water is the gram atomic mass of oxygen plus twice the gram atomic mass of hydrogen = 15.999 + 2(1.008) = 18.015. The balanced reaction equation shows that four molecules of water are produced for each molecule of propane, so that the ratio of grams of propane reacted to grams of water produced is 44.097/(4 X 18.015) = 0.61191, to the justified number of significant digits. Therefore, the amount of propane required to produce 8.00 grams of water = 8.00(0.61191) = 4.90 grams to the clearly justified number of significant digits, matching the three significant digits given for 8.00, or 4.895 grams, where the last digit is depressed because it may well not be accurate within + 1
If all of the quantities stated actually reacted, the law of the conservation of mass shows that the mass of carbon dioxide produced would be 40 - 18 or 22 grams.
18 grams
First you need to write a balanced equation. You are given that propane undergoes a combustion reaction that produces carbon dioxide and water.Unbalanced: C3H8 + O2 ---> CO2 + H2OBalanced: C3H8 + 5O2 ---> 3CO2 + 4H2OGivens:42.0 grams C3H8 (Molecular mass 44.0 g)115.0 grams O2 (Molecular mass 32.0 g)Molecular mass of CO2: 44.0 gMole ratio 1:5:3:4 (C3H8:O2:CO2:H2O)Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.42.0 g C3H8 / (44.0 g) = .955 moles C3H8 115.0 g
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
Assuming that the water is produced by complete combustion of propane, the balanced equation is: C3H8 + 5 O2 = 3 CO2 + 4 H2O. The gram molecular mass of propane is three times the gram atomic mass of carbon plus eight times the gram atomic mass of hydrogen = 3(12.011) + 8(1.008) = 44.097, and the gram molecular mass of water is the gram atomic mass of oxygen plus twice the gram atomic mass of hydrogen = 15.999 + 2(1.008) = 18.015. The balanced reaction equation shows that four molecules of water are produced for each molecule of propane, so that the ratio of grams of propane reacted to grams of water produced is 44.097/(4 X 18.015) = 0.61191, to the justified number of significant digits. Therefore, the amount of propane required to produce 8.00 grams of water = 8.00(0.61191) = 4.90 grams to the clearly justified number of significant digits, matching the three significant digits given for 8.00, or 4.895 grams, where the last digit is depressed because it may well not be accurate within + 1
44
If all of the quantities stated actually reacted, the law of the conservation of mass shows that the mass of carbon dioxide produced would be 40 - 18 or 22 grams.
The reaction isC3H8 + 5O2 ----> 3CO2 + 4H2O100g of propane is approx 2.27 moles.From the equation above, we see that the ratio of C3H8 to CO2 is 1:3, therefore the number of moles of CO2 which form is approx 6.82.This relates to a mass of 300g of CO2
18 grams
First you need to write a balanced equation. You are given that propane undergoes a combustion reaction that produces carbon dioxide and water.Unbalanced: C3H8 + O2 ---> CO2 + H2OBalanced: C3H8 + 5O2 ---> 3CO2 + 4H2OGivens:42.0 grams C3H8 (Molecular mass 44.0 g)115.0 grams O2 (Molecular mass 32.0 g)Molecular mass of CO2: 44.0 gMole ratio 1:5:3:4 (C3H8:O2:CO2:H2O)Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.42.0 g C3H8 / (44.0 g) = .955 moles C3H8 115.0 g
67.5 grams of H2O
21.6 grams
The mass of carbon dioxide is 141,2 g.
Propane has a density of .0018794 g/cm3. in 22.4 liters, there are 22400 cm3, so there are 42.10 grams of propane. The molar mass of propane is 44.10 g/mole, so there are .955 moles. Via Avogadro's number, there are then 5.751 E23 molecules of propane present.
16.9