Answer:
<em>The answer that is true is </em><em>A</em>
Explanation:
The first thing to do is to perform stoichiometry relationships and calculations.
<em>First relationship</em><em>: 0.3 moles of CO₂ are produced
</em>
6 mol CO₂ _____ 1 mol C₆H₁₂O₆
0.3 mol CO₂ _____ X = <em>0.05 mol of C₆H₁₂O₆</em>
Calculation: 0.3 mol x 1 mol / 6 mol = 0.05 mol of C₆H₁₂O₆
<em>Second relationship:</em><em> 0.60 moles of CO₂ are produced
</em>
6 mol CO₂ _____ 1 mol C₆H₁₂O₆
0.6 mol CO₂_____ X = <em>0.1 mol of C₆H₁₂O₆
</em>
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 =<em> 0.6 mol of O₂</em>
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 <em>Proust's law</em> or<em> law of defined proportions</em>, stated as: “<u><em>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</em></u>”. 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 <em>law of defined proportions</em> or <em>Proust's law.</em> 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.