Question

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

Answer 1

Answer:

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

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

$moles of propane=\frac{mass of propane}{mass molar of propane}\\ moles of propane=\frac{100 g}{44.1 g/mol}\\ moles of propane=2.26 mol$

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

$Mass of CO2=Moles of CO2x Molar mass of CO2\\Mass of CO2=6.80 mol x\frac{44.01g}{mol}\\ Mass of CO2=27.3g$

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