The combustion of 0.295 kg of propane produces 712 g of carbon dioxide. What is the percent yield of carbon dioxide? ( Make sure to balance equation) C3H8 (g)+ 029) à co2 g H200 0 a. 124%
b. 41 .4%
c. 80.5%
d. 0.805 %

Option C. 80.5%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Next, we shall determine the mass of C3H8 that reacted and the mass of CO2 produced from the balanced equation.

This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44 g/mol

Mass of C3H8 from the balanced equation = 1 x 44 = 44 g

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Mass of CO2 from the balanced equation = 3 x 44 = 132 g

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Next, we shall determine the theoretical yield of CO2.

This can be obtained as shown below:

From the balanced equation above,

44 g of C3H8 reacted to produce 132 g of CO2.

Therefore, 0.295 kg (i.e 295 g) will react to produce = (295 x 132)/44 = 885 g of CO2.

Therefore, the theoretical yield of CO2 is 885 g.

Finally, we shall determine the percentage yield of CO2 as follow:

Actual yield of CO2 = 712 g

Theoretical yield of CO2 = 885 g

Percentage yield of CO2 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield of CO2 = 712/885 x 100

Percentage yield = 80.5%

Therefore, the percentage yield of CO2 is 80.5%.

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

i think its 5049.

i hope this !