28.76 g of O2
Write the complete combustion reaction and balance
C3H8 + 5O2 yields 3CO2 + 4H2O
set up ratio based on molar masses (stat organized)
44g = 160g
X 7.91g
Solve for X
I have found this method easier for students (gen. chem) as compared to dimensional analysis
16.9
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
the formula is no. moles is mass / molecular mass. As the number of moles is 1, the mass required will be exactly the same as the molecular mass, which is 58.32g
to find molar mass you add the molar mass of the carbons 3(amu)+ molar mass of the hydrogens 8(amu) to find molar mass you add the molar mass of the carbons 3(amu)+ molar mass of the hydrogens 8(amu)
Heat capacity is the amount of heat energy required to raise the temperature of a unit mass of a material one degree.
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?
The mass of CO2 is 1 105 g.
16.9
The answer is 24,15 g.
The mass of carbon dioxide is 878 g.
CO2 + Ca(OH)2 write a balanced equation for the combustion of benzene (C6H6)and find the mass of CO2 produced from the complete combustion of 25g benzene
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
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
Oxygen is NOT a PRODUCT (it is not produced) from the complete combustion of methane, it is a REACTANT (it is used in the reaction). The answer is therefore a mass of zero.
The best example of the law of conservation of matter is complete combustion. If you were to burn something of known mass in a closed system, the system would have the same mass before and after combustion occurs.
density
Carbon dioxide is CO2 and has a molar mass of 44g whereas propane is C3H8 and has a molar mass of 44g as well. Therefore, both are of equal mass