Answer:
13.7 g of CO₂
Limiting reactant: C₆H₁₂O₆
3.81 g of O₂
Step-by-step explanation:
We convert the mass of the reactants to moles, in order to find out the limiting reactant and the excess reagent
9.30 g / 180 g/mol = 0.052 moles of glucose
13.8 g / 32 g/mol = 0.431 moles of oxygen
The equation is: C₆H₁₂O₆(s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)
Ratio is 1:6. Let's consider this rule of three:
1 mol of glucose reacts with 6 moles of oxygen
Then, 0.052 moles of glucose must react with (0.052 . 6) /1 = 0.312 moles
We have 0.431 moles of oxygen and we only need 0.312 moles. This means that an amount of oxygen still remains after the reaction is complete:
0.431 - 0.312 = 0.119 moles. We convert the moles to mass:
0.119 mol . 32 g / 1mol = 3.81 g
In conclussion, the limiting reactant is the glucose.
6 moles of oxygen react with 1 mol of glucose
0.431 moles of O₂ will react with (0.431 . 1) /6 = 0.072 moles of glucose
We only have 0.052 moles, so it is ok to say, that glucose is the limiting cause we do not have enough glucose.
Let's verify, the maximum amount of carbon dioxide that can be formed:
1 mol of glucose can produce 6 moles of CO₂
Therefore 0.052 moles of glucose will produce (0.052 . 6) /1 = 0.312 moles
We convert the moles to mass → 0.312 mol . 44 g /1 mol = 13.7 g