Question

For the following reaction, 3.39 grams of oxygen gas are mixed with excess benzene (C₆H₆) . assume that the percent yield of carbon dioxide is 78.0 %. benzene (C₆H₆)(l) oxygen(g) carbon dioxide(g) water(g) what is the ideal yield of carbon dioxide ?

Answer 1

The ideal yield of carbon dioxide from 3.39 g of oxygen reacting with excess benzene, assuming 100% yield, is 18.64 g of CO2.

Step-by-step explanation:

To calculate the ideal yield of carbon dioxide in this reaction, we need to first determine the amount of carbon dioxide that would be produced if the reaction goes to completion. From the balanced chemical equation, we can see that the mole ratio between benzene and carbon dioxide is 1:3. Therefore, if we know the moles of benzene, we can calculate the moles of carbon dioxide. We can then convert the moles of carbon dioxide to grams using the molar mass of carbon dioxide.

To determine the ideal yield of carbon dioxide (CO2) from the combustion of benzene (C6H6), we must first calculate the moles of oxygen (O2) based on the given mass. Since the molar mass of O2 is 32.00 g/mol, we have 3.39 g / 32.00 g/mol = 0.1059 mol of O2. The combustion reaction of benzene is:

C6H6 + 15/2 O2 → 6 CO2 + 3 H2O

Based on stoichiometry, 1 mol of benzene produces 6 mol of CO2. Oxygen gas is in excess, so all 0.1059 mol of O2 will react, producing 0.1059 / (15/2) * 6 = 0.4236 mol of CO2. The molar mass of CO2 is 44.01 g/mol, hence the ideal yield of CO2 is 0.4236 mol * 44.01 g/mol = 18.64 g of CO2.