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C.H2Os(aq) + 2C,H,OH(aq) +200 (9) Fermentation of 862 ml grape juice (density is 1.0 g/cm') is allowed to take place in a bottle with a total volume of 954 mL until 16% by volume is ethanol (C,HSOH). Assuming that CO, obeys Henry's law, calculate the partial pressure of CO2 in the gas phase and the solubility of CO, in the wine at 25C. The Henry's law constant for CO2 is 3.1 x 10-mol/L atm at 25C with Henry's law in the form C = kP, where is the concentration of the gas in mol/L. (The density of ethanol is 0.79 g/cm) (Enter your answers to two significant figures.) Partial pressure - 640 atm Solubility = 2.4 M 16% 862 mh total lol 954mL vol of ethanol = 16% = 100 x 862mL of Grape juice = 137.92 mL MASS of Ethanol sena vol 137.92mL .793 (cm3 (lon) 108.9568 9 108.9568 = 2.365 x 2.4moles. 46.07 954-862 =92 0.092 L 2.4 mol X .6821L-aton onolik) x2986 092 2.4 x. 0821x298 092 = 638.238 x 640 ATM