Question

The combustion of 100.0 grams of propane (C3H8) produces what mass of carbon dioxide? ___C3H8 + ___O2 ___CO2 + ___H2O

Answer 1

The combustion of 100.0 grams of propane (C3H8) produces 299.24 grams of carbon dioxide (CO2), calculated using stoichiometry based on the balanced chemical reaction of propane with oxygen.

Step-by-step explanation:

The combustion of 100.0 grams of propane (C3H8) results in the production of carbon dioxide (CO2). To find out the mass of carbon dioxide produced, we must first write and balance the chemical equation for the complete combustion of propane.

The balanced chemical equation for the combustion of propane is:

C3H8 + 5O2 → 3CO2 + 4H2O

From this equation, we see that one mole of propane (C3H8) produces three moles of carbon dioxide (CO2). By using the molar mass of propane (44.1 g/mol) and carbon dioxide (44.0 g/mol), we can calculate the mass of CO2 produced.

First, we convert 100.0 grams of propane to moles:

100.0 g C3H8 × (1 mol C3H8 / 44.1 g C3H8) = 2.267 moles C3H8

Then we use the stoichiometric ratio from the balanced equation to find the moles of CO2 produced:

2.267 moles C3H8 × (3 moles CO2 / 1 mol C3H8) = 6.801 moles CO2

Finally, we convert moles of CO2 to grams:

6.801 moles CO2 × (44.0 g CO2 / 1 mol CO2) = 299.24 grams CO2

Therefore, the combustion of 100.0 grams of propane produces 299.24 grams of carbon dioxide.

Answer 2

So, the combustion of 100.0g of propane produces 27.3 g of carbon dioxide.

Step-by-step explanation:

First, what we will do is balance the equation so that the atoms in the reactants match the atoms in the products.

We start by balancing the carbon. We have 3 carbon atoms on the reactant side, so we will place the coefficient 3 on the reactant side of the carbon dioxide molecule.

`C_(3)H_(8)+xO_(2) 3CO_(2)+yH_(2)O`

For hydrogen we have 8 atoms in the reactants, we place the coefficient 4 in front of the water molecule to match.

`C_(3)H_(8)+5O_(2) 3CO_(2)+4H_(2)O`

Finally, we have 10 oxygen in the products. To match the number of oxygen atoms we will place the coefficient 5 in front of the O2 molecule.

`C_(3)H_(8)+5O_(2) 3CO_(2)+4H_(2)O`

So, for each molecule of propane, we will obtain 3 molecules of carbon dioxide (CO2). We will calculate now the number of moles of propane. We will use the mass molar of propane.

Mass molar of propane = 44.1 g/mol

`moles of propane=(mass of propane)/(mass molar of propane)\\ moles of propane=(100 g)/(44.1 g/mol)\\ moles of propane=2.26 mol`

So, we will multiply the moles of propane by 3 to obtain the moles of dioxide carbon.

Moles of CO2 = 2.26 mol x 3 = 6.80 mol of CO2

We already we the moles of CO2, now to convert these moles in mass we will use the molar mass of the CO2.

Molar mass of CO2 = 44.01 g/mol

`Mass of CO2=Moles of CO2x Molar mass of CO2\\Mass of CO2=6.80 mol x(44.01g)/(mol)\\ Mass of CO2=27.3g`

So, the combustion of 100.0g of propane produces 27.3 g of carbon dioxide.