prelab chem 108

Question

For the balanced equation shown below, how many grams of C2H3O2Cl will react with 14.5 grams of HCl? 

2C₂H₃O₂Cl(g) + 3O₂ (g) -->4CO₂ (g) + 2H₂O(g) + 2HCl (g)

Referring to question 3, if a chemist was able to produce 8.59 grams of HCl using 31.2 grams of C₂H₃O₂Cl with excess O₂ and other altering conditions for the same reaction given in question 3, what is the chemist percent yield of HCl?

Answer 1

2C₂H₃O₂Cl(g) + 3O₂ (g) ==> 4CO₂ (g) + 2H₂O(g) + 2HCl (g) ... balanced equation

For the first question, no C₂H₃O₂Cl will react with HCl since HCl is NOT a reactant in the balanced equation.

For the second question:

Theoretical yield HCl = 31.2 g C₂H₃O₂Cl x 1 mol C₂H₃O₂Cl / 94.5 g x 2 mol HCl / 2 mol C₂H₃O₂Cl x 36.5 g HCl / mol = 12.1 g HCl

Percent yield of HCl:

% yield = actual yield / theoretical yield (x100%)

% yield HCl = 8.59 g / 12.1 g (x100%) = 71.3% yield