Answer: 67.2 moles
Explanation: [tex]2C_2H5+5O_2\rightarrow 4CO_2+2H_2O[/tex]
According to the given balanced equation, 2 moles of acetylene [tex]C_2H_2[/tex] combine with 5 moles of oxygen [tex]O_2[/tex] to produce 4 moles of carbon dioxide [tex]CO_2[/tex].
Thus if 2 moles of acetylene [tex]C_2H_2[/tex] combine with = 5 moles of oxygen [tex]O_2[/tex]
35 moles of acetylene [tex]C_2H_2[/tex] combine with=[tex]\frac{5}{2}\times {35}=87.5[/tex] moles of oxygen [tex]O_2[/tex]
But as only 84 moles of oxygen are available, acetylene is not a limiting reagent.
5 moles of oxygen [tex]O_2[/tex] reacts with = 2 moles of acetylene [tex]C_2H_2[/tex]
84 moles of oxygen [tex]O_2[/tex] reacts with=[tex]\frac{2}{5}\times {84}=33.6[/tex] moles of acetylene [tex]C_2H_2[/tex]
Thus Oxygen is the limiting reagent as it limits the formation of products. Acetylene is excess reagent as it is present in excess.
2 moles of acetylene [tex]C_2H_2[/tex] produce= 4 moles of carbon dioxide [tex]CO_2[/tex].
33.6 moles of acetylene [tex]C_2H_2[/tex] produce=[tex]\frac{4}{2}\times {33.6}=67.2[/tex] moles of of carbon dioxide [tex]CO_2[/tex].
Mole is the amount of substance that is equal to Avogadro's constant. 67.2 mole of Carbon dioxide produced when 35.0 mol acetone react completely.
Mole is the amount of substance that is equal to Avogadro's constant.
Given here,
[tex]\bold{ 2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O}[/tex]
Means 2 moles of acetylene react with 5 moles of oxygen in the combustion reaction. to produce 4 moles of Carbon dioxide.
The molar ratio of Oxygen and acetylene is 5:2
The molar ratio of Acetylene and Carbon dioxide = 2:4
So, 84 moles of [tex]\bold { O_2}[/tex], react with
[tex]\bold{\Rightarrow \frac{2}{5} \times 84 = 33.6 mole }[/tex] of acetone
hence,
33.6g of acetone react with
[tex]\bold{\Rightarrow \frac{4}{2} \times 33.6 = 67.2}[/tex] of Carbon dioxide
Therefore, we can conclude that 67.2 mole of Carbon dioxide produced when 35.0 mol acetone react completely.
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