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How many grams of o2 are needed to completely burn 96.6 grams of c3h8?
Wiki User
∙ 10y ago
CO2 has 1 carbon atom and 2 oxygen atoms. Carbon has an atomic mass of about 12 and oxygen 16. The total for carbon dioxide is 12 + 16 + 16 = 44. The make 88gms you would need the same proportion in grams as the proportion of atomic masses. The answer is therefore 64 grams of oxygen. For each tonne of carbon burnt the oxygen from 15000 cubic metres of atmosphere is needed.
Wiki User
∙ 12y ago
Wiki User
∙ 9y ago85g of oxygen is
85/16 = 5.3125 moles of atoms.
There are 2 moles of O atoms per mole of carbon dioxide, so there are 2.65625 moles of carbon dioxide.
This is 2.65625 x 44 g = 116.875 g
Wiki User
∙ 10y agoThe chemical equation for this reaction is:
C3H8(g) + 5O2(g) (arrow) 4H2O(l) + 3CO2(g)
Using stoichiometry, 5 moles of oxygen are needed to burn 1 mole of C3H8. 96.6 grams is equal to 0.5889 moles. You multiply this by five to find the amount of moles of O2 required, which is 2.945 moles. Converting this to grams gives 94.24 grams of O2.
Wiki User
∙ 10y ago85.6 g C * 1 mol C/12 g C * 1 mol O2/1 mol C * 32 g O2/1 mol O2 = 228.27 g O2
Just set it up like this so you can do unit conversion and ta dum! You get the answer you're looking for!
Wiki User
∙ 12y agoC+O2->CO2
proportion: 84,9g(C)- Xg (O2)
12g (MrC)- 32(MrO2)
84,9*32=12X
X=226,4g (O2)
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How many moles of oxygen are necessary to react completely with 4 moles of propane?
4 C3H8 + 20 O2 ---> 12 CO2 + 16 H2O So 20 mol of O2 are needed.
How many grams carbon dioxide is made when 42 grams of propane is burned with 115 grams of oxygen how many grams of water is made?
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
How many grams of CO2 will be produced by burning 34.3 gram of C3H8?
[ 34.3(gC3H8) / 44(g/mol C3H8)] * [ 3mol CO2 / 1molC3H8 ] * 44(g/mol CO2) = 103 gram CO2
When 2.13 moles of C3H8 burn in excess oxygen how many moles of CO2 will be formed assuming complete combustion?
Out of one mole C3H8 three moles carbondioxide (CO2) are formed by complete combustion of it.So 2.13 mole C3H8 make 3*2.13 mole CO2 = 6.39 moleCO2
What is the total number of moles of O2(g) required for the complete combustion of 1.5 moles of C3H8(g)?
C3H8 + 5O2 ==> 3CO2 + 4H2O balanced equationmole ratio O2:C2H8 = 5:1 1.5 moles C3H8 x 5 moles O2/mole C3H8 = 7.5 moles O2 needed
How many grams of 02(g) are needed to completely burn 23.7 g of C38H(g)?
The balanced equation is C3H8 + 5O2 ---> 3CO2 + 4H2O moles C3H8 = 23.7 g x 1 mol/44 g = 0.539 moles moles O2 needed = 5 x 0.539 moles = 2.695 moles O2 (it takes 5 moles O2 per mole C3H8) grams O2 needed = 2.695 moles x 32 g/mole = 86.2 grams O2 needed (3 sig figs)
How many moles of C3H8 are needed to react with 0.567 mol of 02?
You have to burn C3H8 in O2. You get 3CO2 plus 4H2O. So to burn one mole of C3H8, you need 5 moles of O2. That means you need one fifth of C3H8 as compared to O2. So you need 0.567/5 = 0.1134 moles of C3H8. Hence the answer.
What is the word equation for prophane?
Propane ( C3H8 ) Will burn completely in excess oxygen to form water and carbon dioxide;C3H8 + 5 O2 ---> 3 CO2 + 4 H2O
How many grams of oxygen are required to complete combustion of 2.75 moles of propane?
C3H8 + 5O2 --> 3CO2 + 4H2O 2.75 mole C3H8 (5 moles O2/1 mole C3H8)(32 grams/1 moleO2) = 440 grams oxygen required =====================
If 100.0g C3H8 is burned completely how many grams CO2 would be produced if the only source of carbon is C3H8?
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
How many moles of oxygen are necessary to react completely with 4 moles of propane?
4 C3H8 + 20 O2 ---> 12 CO2 + 16 H2O So 20 mol of O2 are needed.
How many grams carbon dioxide is made when 42 grams of propane is burned with 115 grams of oxygen how many grams of water is made?
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
C3H8 plus 5O2---3CO2 plus 4H2O. How many moles of water form when 4.50 L propane gas C3H8 completely react?
5048+w4g5=23vm32
How much oxygen takes to burn propane?
C3H8 + 5O2 -> 3CO2 + 4H2O That is the complete combustion for Propane.
Is produced when propane (C3H8) is combusted completely?
Combustion of propaneC3H8 + 5O2 ==> 3CO2 + 4H2OWhat is the question?
When 2.13 moles of C3H8 burn in excess oxygen how many moles of CO2 will be formed assuming complete combustion?
Out of one mole C3H8 three moles carbondioxide (CO2) are formed by complete combustion of it.So 2.13 mole C3H8 make 3*2.13 mole CO2 = 6.39 moleCO2
How many grams of CO2 will be produced by burning 34.3 gram of C3H8?
[ 34.3(gC3H8) / 44(g/mol C3H8)] * [ 3mol CO2 / 1molC3H8 ] * 44(g/mol CO2) = 103 gram CO2
