This preview shows page 1. Sign up to view the full content.
Unformatted text preview: --------- 104 Chapter Three Stoichiometry
Next we must take into account the fact that each mole of propane reacts with 5 moles of oxygen. The best way to do this is to use the balanced equation to construct a mole ratio. In this case we want to convert from moles of propane to moles of oxygen. From the balanced equation we see that 5 moles of O2 is required for each mole of C3Hs, so the appropriate ratio is 5 mol O2 I mol C3HS Multiplying the number of moles of C3HS by this factor gives the number of moles of O2 required: 2.18~X1~ 5 mol O2
= lO.9 mol O2 Notice that the mole ratio is set up so that the moles of C3Hs cancel out, and the units that result are moles of O2. Since the original question asked for the mass of oxygen needed to react with 96.1 grams of propane, the lO.9 moles of O2 must be converted to grams. Since the molar mass of O2 is 32.0 g/mol, 32.0 g O2 lO.9 meI-02 X 1 meI-02
= 349 g O 2 Therefore, 349 grams of oxygen is required to burn 96.1 grams of propane. This example can be extended by asking: "What mass of carbon dioxide is produced when 96.1 grams of propane is combusted with oxygen?" In this case we must convert between moles of propane and moles of carbon dioxide. This can be accomplished by looking at the balanced equation, which shows that 3 moles of CO2 is produced for each mole of C3HS reacted. The mole ratio needed to convert from moles of propane to moles of carbon dioxide is 3 mol CO2 1 mol C3HS The conversion is 2.18~ X 3 mol CO2
= 1~ 6.54 mol CO2 Then, using the molar mass of CO2 (44.0 g/mol), we calculate the mass of CO2 produced: 6.54~X1~ 44.0 g CO2
= 288 g CO 2 We will now summarize the sequence of steps needed to carry out stoichiometric calculations. 96.1 g C3H8 1 mol C3H8 44.1 g C3H8 2.18 mol C3H8 3 mol C02 1 mol C3H8 6.54 mol C02 44.0 g C02 1 mol C02 288 g C02 ...
View Full Document