Question #4ba54
1 Answer
The reaction will produce
Explanation:
So, start with the balanced chemical equation
#2"C"_2"H"_3"Cl" + 5"O"_2 -> 4"CO"_2 + 2"H"_2"O" + 2"HCl"#
Notice that you have a
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
#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
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
#"10.0 moles " - " 4.00 moles" = "6.00 moles" -># 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
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"#