C6h12o6+6o2→6co2+6h2o

the reaction between c6h12o6 and o2 is represented by the balanced equation above. in an experiment, 0.30mol of co2 was produced from the reaction of 0.05mol of c6h12o6 with excess o2. the reaction was repeated at the same temperature and in the same container, but this time 0.60mol of co2 was produced. which of the following must be true?

a the initial amount of c6h12o6 in the container must have been 0.10mol

b exactly 0.30mol of c6h12o6 must have reacted because c atoms were conserved.

c exactly 0.40mol of o2 must have reacted because the temperature and container volume are the same.

d more than 0.60mol of o2 must have reacted because it was present in excess.

The answer that is true is A

Explanation:

The first thing to do is to perform stoichiometry relationships and calculations.

First relationship: 0.3 moles of CO₂ are produced

6 mol CO₂ 1 mol C₆H₁₂O₆

0.3 mol CO₂ X = 0.05 mol of C₆H₁₂O₆

Calculation: 0.3 mol x 1 mol / 6 mol = 0.05 mol of C₆H₁₂O₆

Second relationship: 0.60 moles of CO₂ are produced

6 mol CO₂ 1 mol C₆H₁₂O₆

0.6 mol CO₂ X = 0.1 mol of C₆H₁₂O₆

Calculation: 0.6 mol x 1 mol / 6 mol = 0.1 mol of C₆H₁₂O₆

6 mol CO₂ 6 mol O₂

0.6 mol CO₂ X = 0.6 mol of O₂

Calculation: 0.6 mol x 6 mol / 6 mol = 0.6 mol of O₂

Then, the answer that is true is A, because in order for 0.6 moles of CO₂ to be produced, there must initially be 0.1 mol of C6H12O6 as calculated.

The B is not correct because for more CO₂ to be produced, more C₆H₁₂O₆ must react, this is defined according to Proust's law or law of defined proportions, stated as: “When the combination of two or more elements occurs to form a new compound, they always do it in a mass relationship that is completely constant”. It is by this law that no more CO₂ can be produced with the same amount of C₆H₁₂O₆ reagent.

The C is not correct either, although the temperature and volume of the container are the same, the O₂ is in excess to react, so 0.6 mol of it is needed to produce 0.6 mol of CO₂, according to the calculations made .

The D is not correct because it would also not respect the law of defined proportions or Proust's law. Although O₂ is in excess, it cannot react more than the stoichiometric ratio. Simply, if it is in excess, the rest of the O₂ will remain unreacted.

what are the óptions. pléase list them

answer: mercury

explanation: