Question

Please help 3. The combustion of glucose is represented by the following balanced equation: C6H12O6 + 602 - 6H20 + 6CO2.

Which reactant is the limiting reagent if there is 1 gram of both C6H12O6 and O2? (Hint: Must convert to moles to determine this.)
O a C.H 20
O bO2
O c H₂O
Od CO₂

Answer 1

Step-by-step explanation:

C₆H₁₂O₆ + 60₂ = 6H₂0 + 6CO₂.

1 mole 6 mole

molecular weight of C₆H₁₂O₆ = 180 g

molecular weight of oxygen = 32 g

1 gram of glucose = 1 / 180 = 5.55 x 10⁻³ moles

1 gram of oxygen = 1 / 32 = 31.25 x 10⁻³ moles

1 mole of glucose reacts with 6 moles of oxygen

5.55 x 10⁻³ moles of glucose will react with 6 x 5.55 x 10⁻³ moles of oxygen

= 33.30 x 10⁻³ moles of oxygen .

But oxygen available = 31.25 x 10⁻³ moles

So available oxygen is less than required .

Hence oxygen is the limiting reagent .

b ) is the right option .