The mixture of Na2CO3 and NaHCO3 weighs 220g. Heated precipitate weighs 182g. What is the percent composition?

1 Answer
Mar 13, 2016

Answer:

The mixture is #"53.2 % Na"_2"CO"_3# and #"46.8 % NaHCO"_3#.

Explanation:

When you heat a mixture of #"Na"_2"CO"_3# and #"NaHCO"_3#, only the #"NaHCO"_3# decomposes.

The equation for the decomposition is

#"2NaHCO"_3("s") → "Na"_2"CO"_3("s") + "CO"_2("g") + "H"_2"O"("g")#

The loss in mass is caused by the loss of #"H"_2"O"# and #"CO"_2# (equivalent to the loss of #"H"_2"CO"_3#).

The loss in mass is #"(220 – 182) g = 38 g"#.

This corresponds to

#38 color(red)(cancel(color(black)("g H"_2"CO"_3))) × ("1 mol H"_2"CO"_3)/(62.02 color(red)(cancel(color(black)("g H"_2"CO"_3)))) = "0.613 mol H"_2"CO"_3#

#"Moles of NaHCO"_3 = 0.613 color(red)(cancel(color(black)("mol H"_2"CO"_3))) × ("2 mol NaHCO"_3)/(1 color(red)(cancel(color(black)("mol H"_2"CO"_3)))) = "1.225 mol NaHCO"_3#

and

#"Mass of NaHCO"_3 = 1.2225 color(red)(cancel(color(black)("mol NaHCO"_3))) × ("84.01 g NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaHCO"_3)))) = "102.9 g NaHCO"_3#.

The loss of mass after heating corresponds to #"102.9 g of original NaHCO"_3#.

The rest of the original mixture must have been #"Na"_2"CO"_3#.

#"Mass of original Na"_2"CO"_3 = "220 g – 102.9 g" = "117.1 g"#

#"% Na"_2"CO"_3 = (117.1 color(red)(cancel(color(black)("g"))))/(220 color(red)(cancel(color(black)("g")))) × 100 % = "53.2 %"#

#% "NaHCO"_3 = (102.9 color(red)(cancel(color(black)("g"))))/(220 color(red)(cancel(color(black)("g")))) × 100 % = 46.8 %#