Question #e4dc6
1 Answer
Here's what I got.
Explanation:
The important thing to do here is write a balanced chemical equation for this decomposition reaction.
Sodium bicarbonate,
#color(red)(2)"NaHCO"_text(3(s]) -> "Na"_2"CO"_text(3(s]) + "CO"_text(2(g]) + "H"_2"O"_text((g])#
Notice that you have a
Use sodium carbonate's molar amss to determine how many moles you'd get in that sample
#0.685color(red)(cancel(color(black)("g"))) * "1 mole NaHCO"_3/(84.007color(red)(cancel(color(black)("g")))) = "0.008154 moles NaHCO"_3#
Now, if the reaction were to have a
#0.008154color(red)(cancel(color(black)("moles NaHCO"_3))) * ("1 mole Na"_2"CO"_3)/(color(red)(2)color(red)(cancel(color(black)("moles NaHCO"_3)))) = "0.004077 moles Na"_2"CO"_3#
Use the molar mass of sodium carbonate to determine how many grams would contain this many moles
#0.004077color(red)(cancel(color(black)("moles"))) * "105.99 g"/(1color(red)(cancel(color(black)("mole")))) = "0.4321 g Na"_2"CO"_3#
This will be your reaction's theoretical yield, which tells you how much product you can expect when all the moles of the reactant actually form product.
Now, you know that the reaction produced
This tells you that not all the moles of sodium carbonate reacted to produce sodium carbonate. In other words, the reaction did not have a
Percent yield is defined as
#color(blue)("% yield" = "actual yield"/"theoretical yield" xx 100)#
In your case, the reaction had a percent yield of
#"% yield" = (0.418color(red)(cancel(color(black)("g"))))/(0.4321color(red)(cancel(color(black)("g")))) xx 100 = color(green)(96.7%)#