Methanol (CHOH) combusts according to the following chemical reaction. CHOH(I) + 02(g) → COz(g) + H20(1) 10.4 grams of methanol and 16.3 grams of oxygen are mixed. If 10.1 grams of COz are formed, then what is the percent yield for the reaction?

To find the percent yield of the reaction, we need to compare the actual yield to the theoretical yield and then calculate the percentage.

Step 1: Calculate the molar mass of each compound:
- Methanol (CHOH): 12.01 + 1.01 + 16.00 + 1.01 = 30.03 g/mol
- Oxygen (O2): 16.00 * 2 = 32.00 g/mol
- Carbon Dioxide (CO2): 12.01 + 16.00 * 2 = 44.01 g/mol
- Water (H2O): 1.01 * 2 + 16.00 = 18.02 g/mol

Step 2: Calculate the moles of each substance:
- Moles of methanol: 10.4 g / 30.03 g/mol = 0.346 moles
- Moles of oxygen: 16.3 g / 32.00 g/mol = 0.509 moles

Step 3: Determine the limiting reactant:
Using the balanced chemical reaction, we see that methanol has a 1:1 ratio with carbon dioxide. Therefore, methanol is the limiting reactant because there are fewer moles of methanol than oxygen available.

Step 4: Calculate the theoretical yield of CO2:
Since methanol is the limiting reactant, the moles of CO2 formed will be equal to the moles of methanol used.
Moles of CO2 = 0.346 moles

Step 5: Calculate the theoretical yield of CO2 in grams:
Theoretical yield of CO2 = moles of CO2 * molar mass of CO2
Theoretical yield of CO2 = 0.346 moles * 44.01 g/mol = 15.22 g

Step 6: Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100%
Percent yield = (10.1 g / 15.22 g) * 100% = 66.49%

\boxed{66.49\%} yield for the reaction.