Sally and Sam reacted 20.0 mL of 1.0 M silver (I) chloride with metallic zinc in class. Sam and Sally collected 0.85 g of zinc chloride. What is the percent yield?
Start by writing a balanced chemical equation for this single replacement reaction
#"Zn"_text((s]) + color(red)(2)"AgCl"_text((aq]) -> "ZnCl"_text(2(aq]) + 2"Ag"_text((s])#
Notice that you have a
Now, the problem doesn't provide information about the mass of zinc metal used in the experiment, so you can just assume that zinc is In excess.
This means that all the moles of silver chloride will take part in the reaction.
Use the solution's molarity to determine how many moles you have in that sample
#color(blue)(c = n/V implies n = c * V)#
#n_(AgCl) = "1.0 M" * 20.0 * 10^(-3)"L" = "0.020 moles AgCl"#
The reaction should theoretically produce
#0.020 color(red)(cancel(color(black)("moles AgCl"))) * "1 mole ZnCl"_2/(color(red)(2)color(red)(cancel(color(black)("moles AgCl")))) = "0.010 moles ZnCl"_2#
To convert this to grams, use zinc chloride's molar mass
#0.010 color(red)(cancel(color(black)("moles ZnCl"_2))) * "136.286 g"/(1color(red)(cancel(color(black)("mole ZnCl"_2)))) = "1.363 g"#
So, this tells you that the reaction's theoretical yield, which is what you get in the hypothetical case in which a reaction has a
However, you only recover
The percent yield will thus be
#color(blue)("% yield" = "actual yield"/"theoretical yield" xx 100)#
Plug in your values to get
#"% yield" = (0.85color(red)(cancel(color(black)("g"))))/(1.363color(red)(cancel(color(black)("g")))) xx 100 = color(green)("62%")#
The answer must be rounded to two sig figs, since that is how many sig figs you have for the molarity of the silver chloride solution and for the mass of zinc chloride recovered.