Question

Determine the limiting reactant and grams of co2 produced for the combusion of C6H12O6 if 36 g is reacted with 48 g of oxygen

C6H12O6 +O2---->CO2 (?) +H20.

Answer 1

n=mass/molecular mass
n=35/180
0.2 moles of C6H12O6, 1.5 moles of Oxygen
C6H12O6+6O2-->6CO2+6H2O
1.5=mass/44g, mass=44*1.5=66grams of CO2 produced

Answer 2

Hey Abmon after i got the 0.2 and the 1.5 .how come u didnt do this.

Answer 3

c6h12o6 0.2 x 6O2/1 C6H12O6 =1.2 O2

Answer 4

then i did it for oxygen. i got 1.5 moles x 1 c6h12o6/6o2=O.25 OF C6H12O6