# Question 4ba54

Jan 30, 2015

The reaction will produce $\text{4.00 moles}$ fo $\text{H"_2"O}$.

#### Explanation:

$2 \text{C"_2"H"_3"Cl" + 5"O"_2 -> 4"CO"_2 + 2"H"_2"O" + 2"HCl}$

Notice that you have a $2 : 5$ mole ratio between chloroethylene and oxygen; what this means is that every $2$ moles of the chloroethylene will need $5$ moles of oxygen for the reaction to take place.

Right off the bat, you can see that this criterion is not met for the number of moles you start with. As a result, this becomes a limiting reagent problem.

If you start with $\text{10.0}$ moles of chloroethylene, you would need

10.0 cancel("moles C"_2"H"_3"Cl") * ("5 moles O"_2)/(2 cancel("moles C"_2"H"_3"Cl")) = "25.0 moles O"_2

You only have $10.0$ moles of oxygen. What this means is that oxygen will be your limiting reagent, i.e. the amount of oxygen you have will determine how much chloroethylene will react.

You can determine how much chloroethylene will react by going backward

10.0 cancel("moles O"_2) * ("2 moles C"_2"H"_3"Cl")/(5 cancel("moles O"_2) ) = "4.00 moles C"_2"H"_3"Cl"

The rest of the chloroethylene will remain unreacted

$\text{10.0 moles " - " 4.00 moles" = "6.00 moles} \to$do not take part in the reaction

Now look at the mole ratio you have between chloroethylene and water. Every two moles of the former will produce two moles of the latter - this means you have a $1 : 1$ ratio between the two.

Therefore, the number of moles of water your reaction will produce is

4.00 cancel("moles C"_2"H"_3"Cl") * ("1 mole H"_2"O")/(1 cancel("mole C"_2"H"_3"Cl")) = "4.00 moles H"_2"O"#