For the equation C6 H12 O6 = 2C2 H5 OH + 2CO2, how much ethanol is produced in grams if we start with 750 grams of glucose?

Molar mass of C₆H₁₂O₆ = (12.0×6 + 1.0×12 + 16.0×6) g/mol = 180 g/mol
Molar mass of C₂H₅OH = (12.0×2 + 1.0×6 + 16.0) g/mol = 46.0 g/mol

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Mole ratio C₆H₁₂O₆ : C₂H₅OH = 1 : 2

No. of moles of C₆H₁₂O₆ reacted = (750 g) / (180 g/mol) = 25/6 mol
No. of moles of C₂H₅OH produced = (25/6 mol) × 2 = 25/3 mol
Mass of C₂H₅OH produced = (25/3 mol) × (46.0 g/mol) = 383 mol

C6 H12 O6 = 2C2 H5 OH + 2CO2 ?
I think you mean (the extra spaces confuse things)
C₆H₁₂O₆ ➜ 2C₂H₅OH + 2CO₂

atomic mass
C = 12
H = 1
O = 16
C₆H₁₂O₆ = 6•12+12+6•16 = 180
2C₂H₅OH = 2(12•2+5+16+1) = 92
2CO₂ = 2(12+32) = 88
check 180 = 88 + 92

1 mole of C₆H₁₂O₆ ➜ 2 mole of C₂H₅OH + 2 moles of CO₂
180 grams of C₆H₁₂O₆ ➜ 92 grams of C₂H₅OH + 88 grams of CO₂

how much ethanol is produced in grams if we start with 750 grams of glucose

the ratio is
92/180 = x/750
x = 750•92/180 = 383 g