Question

PLEASE HELP

iv. A total of 132.33g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed.



• Determine if one of the reactants is a limiting reagent.








• How many grams of CO2 and H2O will be produced? (2 points)













b. If the furnace is not properly adjusted, the products of combustion can include other gases, such as CO and unburned hydrocarbons. If only 269.34 g of CO2 were formed in the above reaction, what would the percent yield be? (2 points)

Answer 1

See explanation

Step-by-step explanation:

The equation of the reaction is;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Number of moles of C3H8 = 132.33g/44g/mol = 3 moles

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

Number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

Hence C3H8 is the limiting reactant.

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

b) Actual yield = 269.34 g

Theoretical yield = 396 g

% yield = actual yield/theoretical yield × 100/1

% yield = 269.34 g /396 g × 100

% yield = 68%