Percent Yield Question

Combustion of 6.40 g of CH3OH with 10.2 g of O2 produced 6.12g CO2. The reaction occuring is:

2CH3OH + 3O2 --> 2CO2 + 4H20

calculate the percent yield of CO2

i got theoretical = 8.790
actual = 6.12

i got 69.62% yield of CO2

can someone tell me if im right? thank you so much :) !

2 CH3OH + 3 O2 → 2 CO2 + 4 H2O

(6.40 g CH3OH) / (32.04200 g CH3OH/mol) = 0.19974 mol CH3OH
(10.2 g O2) / (31.99886 g O2/mol) = 0.31876 mol O2

0.31876 mole of O2 would react completely with 0.31876 x (2/3) = 0.21251 mole of CH3OH, but there is not that much CH3OH present, so CH3OH is the limiting reactant.

(0.19974 mol CH3OH) x (2 mol CO2 / 2 mol CH3OH) x (44.00964 g CO2/mol) =
8.79 g CO2 in theory

(6.12 g) / (8.79 g) = 0.696 = 69.6% yield

Except maybe for fussy significant digits, you are correct.