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Calculate the amount of carbon dioxide produced when burning 10.0 g of propane with 5 g of oxygen in the following reaction.
C3H8 + 5 O2 -------> 3 CO2 + 4 H2O
Let us first calculate the number of moles of O2 and propane used in the reaction. The molar mass of propane and O2 are, 44.1 g/mol and 32 g/ mol, respectively. The moles of propane and O2 can be calculated using following formula
Moles = weight/ molar mass
Moles of propane = 10 g/ 44.1 g/mol = 0.227 mol
Moles of O2 = 5 g / 32 g/mol = 0.156 mol
From the balanced chemical equation it is clear that 1 mol of propane requires 5 moles of O2. Therefore, we can consider
1 mol propane = 5 mol of O2
For. 0.227 mol propane = 0.227 x 5 = 1.135 mol of O2
Thus for 0.227 mol of propane 1.135 mol of O2 are required. However, actual number of moles of O2 used are 0.156 mol. Therefore, propane is used in excess while O2 is the limiting reagent as it getting consumed completely. Therefore, the yield of carbon dioxide will depend on the number of moles of O2 used in the reaction.
From the balanced chemical equation it is clear that 5 moles of O2 produce 3 moles of CO2. Therefore, we can consider
5 moles of O2 = 3 moles of CO2
For. 0.156 moles O2 = (0.156 x 3) / 5 mol of CO2
= 0.094 moles of CO2
The molar mass of CO2 is 44.01 g/mol. The mass of CO2 produced can be calculated using following formula
Grams of CO2 = moles x molar mass
= 0.094 mol x 44.01 g/mol
= 4.12 grams
Therefore, 4.12 grams of CO2 will be produced
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