Question

Consider the reaction between CH3CHO(l) and O2(g) to form CO2(g) and H2O(l). If the percent yield of CO2(g) is 56.0% and 18.0 grams of CO2(g) forms, determine the theoretical yield of CO2(g) in moles.

Answer 1

Given:

Percent yield = 56.0 %

Actual yield = 18.0 g

Use:

Percent yield = actual yield * 100 / theoretical yield

56.0 = 18.0 * 100 / theoretical yield

theoretical yield = 32.14 g

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass(CO2)= 32.14 g

use:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(32.14 g)/(44.01 g/mol)

= 0.7303 mol

Answer: 0.730 mol