How do you find the mass of CO2 produced in a reaction of 150.0g of C6H12 in sufficient (excess) oxygen if the reaction has a 35.00% yield?

Question

(3) marks are available for this question, so I'm guessing that there are about three steps to this question...?

Caleb Adams · Accepted Answer

The balanced equation of the combustion reaction of #C_6H_12#

#C_6H_12+9O_2->6CO_2+6H_2O#

This formula indicates that

1 mol #C_6H_12# produces 6 mol #CO_2# on reacting with excess oxygen.

Molar mass of #C_6H_12=6xx12+12xx1=84 "g/mol"#

Molar mass of #CO2=12+2xx16=44 "g/mol"#

So theoretically 84g #C_6H_12# produces #6xx44=264 g # #CO_2#

150g #C_6H_12# produces #264/84xx150 g # #CO_2#

However, the actual yield of CO2 is 35%.

The mass #CO_2# practically produced

#=264/84xx150 g xx35%=165g#

Jeremiah Adams · Answer

undefined To find the mass of CO2 produced, calculate the theoretical yield using the balanced chemical equation. Then, multiply the theoretical yield by the given yield percentage.

Balanced equation: C6H12 + 9O2 → 6CO2 + 6H2O

Molar mass of C6H12: 84.18 g/mol
Molar mass of CO2: 44.01 g/mol

Calculate moles of C6H12, then moles of CO2, and finally the mass of CO2:

1. Moles of C6H12 = 150.0 g / 84.18 g/mol
2. Moles of CO2 (using the mole ratio from the balanced equation)
3. Mass of CO2 = Moles of CO2 * Molar mass of CO2

Finally, apply the given yield percentage to find the actual mass of CO2 produced.