How many grams of CO2 will be produced when 29.0g of C3 H8 reacts with extra O2?
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Follow these five steps
(1) Calculate the number of moles of C3H8 that are oxidized.
(2) Write the balanced equation of combustion.
(3) The balanced reaction equation will tell you how many moles of CO2 are produced per mole of C3H8. That number is three.
(4) Calculate the number of moles of CO2 formed.
(5) How many grams of CO2 is that?
134.0g CO2
To determine the number of grams of CO2 produced when 29.0g of C3H8 reacts with excess O2, we need to balance the equation for the reaction between C3H8 (propane) and O2 (oxygen).
The balanced equation for the combustion of propane is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that for every 1 mole of C3H8 that reacts, 3 moles of CO2 are produced.
To find the number of moles of C3H8, we divide the given mass of C3H8 by its molar mass:
Molar mass of C3H8 = (3*12.01 g/mol) + (8*1.01 g/mol) = 44.10 g/mol
Number of moles of C3H8 = 29.0g / 44.10 g/mol ≈ 0.657 mol
Since the stoichiometry of the balanced equation is 1:3 for C3H8 to CO2, the number of moles of CO2 produced will be 3 times the number of moles of C3H8:
Number of moles of CO2 = 0.657 mol * 3 = 1.971 mol
Finally, we can calculate the grams of CO2 produced by multiplying the number of moles of CO2 by its molar mass:
Molar mass of CO2 = (12.01 g/mol) + (2*16.00 g/mol) = 44.01 g/mol
Grams of CO2 = 1.971 mol * 44.01 g/mol ≈ 86.67 g
Therefore, approximately 86.67 grams of CO2 will be produced when 29.0 grams of C3H8 reacts with excess O2.