Answer:
B. O₂
Explanation:
First we convert 192 g of O₂ into moles, using its molar mass:
- 192 g ÷ 32 g/mol = 6 mol O₂
Now we calculate with how many CS₂ moles would 6 moles of O₂ react with, using the stoichiometric coefficients:
- 6 mol O₂ * [tex]\frac{1molCS_2}{3molO_2}[/tex] = 2 mol CS₂
As there are 4.50 moles of CS₂ and only 2 moles would react, CS₂ is in excess and O₂ is the limiting reactant.
The correct answer is option B.
