How many grams of CO2 can be produced when 200 g of C5H12 are reacted with 300 g of O2?
To determine the number of grams of CO2 produced when 200 g of C5H12 (pentane) reacts with 300 g of O2 (oxygen), we need to calculate the balanced equation and perform a stoichiometric calculation.
First, let's obtain the balanced equation for the combustion of pentane, C5H12, in the presence of oxygen, O2:
C5H12 + O2 -> CO2 + H2O
The balanced equation indicates that one mole of C5H12 reacts with eight moles of O2 to produce five moles of CO2 and six moles of H2O.
Next, we need to calculate the number of moles for both C5H12 and O2, given their respective masses. To do this, we'll use the molar mass of each compound.
The molar mass of C5H12 is 72.15 g/mol (5 * 12.01 g/mol + 12 * 1.01 g/mol).
The molar mass of O2 is 32.00 g/mol (2 * 16.00 g/mol).
Now, we can calculate the number of moles for each compound:
Number of moles of C5H12 = Mass of C5H12 / Molar mass of C5H12
= 200 g / 72.15 g/mol
= 2.772 mol (rounded to three decimal places)
Number of moles of O2 = Mass of O2 / Molar mass of O2
= 300 g / 32.00 g/mol
= 9.375 mol (rounded to three decimal places)
From the balanced equation, we know that the mole ratio between C5H12 and CO2 is 1:5. Therefore, 2.772 moles of C5H12 will produce 5 * 2.772 = 13.86 moles of CO2.
Finally, to calculate the mass of CO2 produced, we can multiply the number of moles of CO2 by its molar mass:
Mass of CO2 = Number of moles of CO2 * Molar mass of CO2
= 13.86 mol * 44.01 g/mol
= 610.32 g (rounded to two decimal places)
Hence, when 200 g of C5H12 reacts with 300 g of O2, approximately 610.32 grams of CO2 will be produced.