Question

2.21 g of ethanol C2H5OH and 3.55 of oxygen O2 are reacted. Molar mass of C2H5Oh 46.08 g/mol. O2 32.00 g/mol. Co2 44.01 g/mol. H20 18.02 g/mol

a)What is the limiting reactant?

b) what is the theoretical yield of carbon dioxide in grams will be obtained in the combustion reaction. starting with the amount of reagents given in part (a). assuming the reaction goes in 100% yield.

c) If 0.951 f of CO2 is obtained from the reaction above, what is the percent yield

D) How many ethanol (in grams) is left over in the reaction of part b. assuming no other side products formed during the reaction?

Answer 1

Answer – We are given,

Mass of Ethanol, C₂H₅OH = 2.21 g

Mass of Oxygen, O₂ = 3.55 g

Actual yield of CO₂ = 0.951 g

The combustion reaction of ethanol is,

C₂H₅OH +3 O₂ -----> 2 CO₂ + 3 H₂O

a) The limiting reactant

To determine the limiting reactant we need to first calculate moles of each reactant and then moles of any one product from both reactant.

We know the formula for moles from mass,

$QHRoDLwhXdY+QAAAABJRU5ErkJggg==$
By plugging the values,

$grJDqYWlkAAAAASUVORK5CYII=$

The moles of CO₂ from ethanol –

From the balanced reaction –

1 mole of C₂H₅OH = 2 moles of CO₂

So, 0.0480 moles of C₂H₅OH =? moles of CO₂

$AHk4kabdRRCjwAAAABJRU5ErkJggg==$

The moles of CO₂ from Oxygen –

From the balanced reaction –

From the balanced reaction –

3 moles of O₂ = 2 moles of CO₂

So, 0.111 moles of O₂ =? moles of CO₂

$0Jwuy6c4fgDphCwPC+64AAAAASUVORK5CYII=$

Thus, the moles of CO₂ are lowest from the reactant oxygen (O₂), hence the reactant oxygen is limiting reactant. The moles of CO₂ are 0.0739 moles.

b) The Theoretical Yield of CO₂

We know the formula for calculating the mass from the mole,

We calculated the moles of CO₂ and it is 0.0739 moles. By plugging the values,

$wNXU1LWniv2rQAAAABJRU5ErkJggg==$

Thus, the theoretical yield of CO₂ is 3.25 g.

c) The Percent Yield of CO₂ –

We know the formula,

$gdwLo59Jw1tPAAAAABJRU5ErkJggg==$

Thus, the percent yield of CO₂ is 29.2 %.

d) The Mass of Excess Reactant –

The excess reactant is ethanol. The mass of ethanol left over is the difference between the mass of ethanol initially given and after it is used in the reaction.

The mass of ethanol used in a reaction –

From the balanced reaction –

1 mole of O₂ = 3 moles of C₂H₅OH

So, 0.111 moles of O₂ =? Moles of C₂H₅OH
$AWSshHhVUjyaAAAAABJRU5ErkJggg==$

The moles of ethanol used in the reaction are 0.0369 moles.

The moles o ethanol remaining after the reaction = 0.0480 – 0.0369 = 0.0110 moles

The mass of ethanol –

$gcUN8QL8zVoawAAAABJRU5ErkJggg==$

Thus, 0.506 gram of ethanol is left over in the reaction.