Question #52580
1 Answer
Explanation:
The first thing that you need to do here is to figure out how many grams of magnesium chloride would be produced by the reaction at
To do that, use the molar mass of magnesium chloride
#3 color(red)(cancel(color(black)("moles MgCl"_2))) * "95.211 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = "285.63 g"#
This means that at
However, you know that the reaction has a
You can thus say that your reaction will produce
#285.63 color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory"))) * overbrace(("73 g MgCl"_2 color(white)(.)"produced")/(100color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory")))))^(color(blue)("= 73% yield"))#
# = color(darkgreen)(ul(color(black)("200 g MgCl"_2color(white)(.)"produced")))#
The answer is rounded to one significant figure, the number of sig figs you have for the number of moles of magnesium chloride that would be produced at