For the reaction C6H 1206C02+ H208 428 g of C6 H 1206 is consumed producing a 3.1 M solution of CO2 what is the volume of CO2 produced

I think it’s m(C6H12O6) = 856 g

M(C6H12O6) = 12*6+1*12+16*6 = 180 g/mol

n(C6H12O6) = m/M = 856 g / 180 g/mol = 4.756 mol

n(CO2) = 6*4.756 = 28.536 mol

M(CO2) = 12+16*2 = 44 g/mol

m(CO2) = n*M = 28.536mol*44g/mol = 1255.584 g