Question #45b1a

1 Answer
May 28, 2017

Answer:

#X = "none of the above"#; #X = 1.2 × 10^23color(white)(l) "molecules of NH"_3#

Explanation:

Step1. Write the half-reaction for the oxidation of dichloroaceticacid

#"CHCl"_2"COOH" +2"H"_2"O" → "2CO"_2 + "Cl"_2 +"6H"^"+" + "6e"^"-"#

Step 2. Calculate the amount of dichloroacetic acid

Six electrons are transferred in the half-reaction, so

#"1 eq" = 1/6 "mol"#

#1.2 color(red)(cancel(color(black)("eq"))) × (1/6 "mol")/(1 color(red)(cancel(color(black)("eq")))) = "0.20 mol"#

Step 3. Calculate the molecules of #"NH"_3# neutralized

#"CHCl"_2"COOH" + "NH"_3 → "CHCl"_2"COO"^"-"color(white)(l) "NH"_4^"+"#

#"Molecules of NH"_3#

#= 0.20 color(red)(cancel(color(black)("mol CHCl"_2"COOH"))) × (1 color(red)(cancel(color(black)("mol NH"_3))))/(1 color(red)(cancel(color(black)("mol CHCl"_2"COOH")))) × (6.022 ×10^23 "molecules NH"_3)/(1 color(red)(cancel(color(black)("mol NH"_3)))) = 1.2 × 10^23 color(white)(l)"molecules NH"_3#