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Q: The number of moles of propane C3H8 in 88 grams of propane?

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0.25 mol

1 mole of propane weight 44 g. So two moles will weigh 88 g.

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

The complete combustion of propane occurs according to the following balanced equation: C3H8 + 5 O2 -> 3 CO2 + 4 H2O. This equation shows that three moles of carbon dioxide are produced for each mole of propane burned. Therefore, burning 7.5 moles of propane will produce 22.5 moles of carbon dioxide. The gram molecular mass of carbon dioxide is 44.01; therefore, 22.5 moles has a mass of 22.5 x 44.01 or 9.9 X 102 grams, to the justified number of significant digits.

The reaction for the complete combustion of propane is: C3H8 + 5O2 --> 3CO2 + 4H2O So for every mole of propane combusted 3 moles of carbon dioxide are formed. Therefore combusting 10 moles of propane will form 30 moles of carbon dioxide.

Related questions

C3H8 + 5O2 --> 3CO2 + 4H2O 2.75 mole C3H8 (5 moles O2/1 mole C3H8)(32 grams/1 moleO2) = 440 grams oxygen required =====================

0.25 mol

Propane is C3H8 and the combustion equation is C3H8 + 5O2 ==> 3CO2 + 4H2OSo the complete combustion of 1 mole of propane requires 5 moles of oxygen.

The chemical name for molecular formula C3H8 is propane. Propane has a weight of 44.10 grams per mole. Carbon (C) has a weight of 12.01 grams per mole. 1 mole of propane contains 3 moles of carbon. Therefore 81.70 percent by mass of propane is C.

Since propane has the formula C3H8, each mole of propane will have 8 moles of hydrogen atoms, so 5 moles of propane will contain 5x8=40 moles of hydrogen.

C3H8 + 5O2 à 3CO2 + 4H2O So 5 moles of oxygen are needed for each mole of propane. The answer is 12.5 moles.

1 mole of propane weight 44 g. So two moles will weigh 88 g.

The volume of propane is 48,93 L.

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

The complete combustion of propane occurs according to the following balanced equation: C3H8 + 5 O2 -> 3 CO2 + 4 H2O. This equation shows that three moles of carbon dioxide are produced for each mole of propane burned. Therefore, burning 7.5 moles of propane will produce 22.5 moles of carbon dioxide. The gram molecular mass of carbon dioxide is 44.01; therefore, 22.5 moles has a mass of 22.5 x 44.01 or 9.9 X 102 grams, to the justified number of significant digits.

The reaction for the complete combustion of propane is: C3H8 + 5O2 --> 3CO2 + 4H2O So for every mole of propane combusted 3 moles of carbon dioxide are formed. Therefore combusting 10 moles of propane will form 30 moles of carbon dioxide.

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.

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