Answer: 3.52 grams
Explanation:
[tex]2C_6H_6+15O_2\rightarrow 12CO_2+6H_2O[/tex]
Given moles:
moles of oxygen = 6.62
moles of [tex]C_6H_6=\frac{\text {given mass}}{\text {molar mas}}=\frac{6.59g}{78g/mol}=0.08moles[/tex]
According to stoichiometry:
2 moles of benzene reacts with = 15 moles of oxygen
Thus 0.08 moles of benzene will react with = [tex]\frac{15}{2}\times 0.08=0.6[/tex] moles of oxygen
Thus benzene is the limiting reagent as it limits the formation of products and oxygen is the excess reagent.
As 2 moles of benzene produce = 12 moles of carbon dioxide
0.08 mole of benzene will produce =[tex]\frac{12}{2}\times 0.08=0.48moles[/tex] of carbon dioxide
Mass of carbon dioxide=[tex]moles\times {\text {Molar mass}}=0.48\times 44=3.52g[/tex]
Thus 3.52 g of carbon dioxide are produced.
