Question #bfd77

2 Answers
May 24, 2016

Answer:

#74.8081%#, rounded to four decimal places

Explanation:

Let the original mixture has no. of grams of Sodium carbonate #=x#
#:.# Number of grams of sodium hydrogen carbonate#=100-x#

The given reaction is

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

From Mole concept we see that
#2# Moles of sodium hydrogen carbonate on heating give #1# mole of sodium carbonate.

Average Molar mass of #"NaHCO"_3# #=22.9898+1.0079+12.011+3xx15.9994=84.0069#
#:.# Average Molar mass of #2# moles of #"NaHCO"_3# #=2xx84.0069=168.0138#

Average Molar mass of #"Na"_2"CO"_3# #=2xx22.9898+12.011+3xx15.9994=105.9888#

#168.0138# gm moles of sodium hydrogen carbonate on heating give gm moles of sodium carbonate #=105.9888#
#100-x# -do- #=105.9888/168.0138xx(100-x)#
Total sodium carbonate in the final mixture#=105.9888/168.0138xx(100-x)+x#, Given #=90.7#
#=>105.9888/168.0138xx(100-x)+x=90.7#
Solving for #x#, we get
#x=74.8081#, rounded to four decimal places
As the original mixtuer is #100# #g#
Percentage of sodium carbonate#=74.8081%#, rounded to four decimal places

May 24, 2016

Answer:

74.8%

Explanation:

On heating the given mixture of 100g of #NaHCO_3 and Na_2CO_3# ,only# NaHCO_3# will decompose according to the following balanced equation but other component #Na_2CO_3# will not.

The Equation

#2NaHCO_3(s)->Na_2CO_3(s)+CO_2(g)+H_2O(g)#

According to this balanced equation 2 moles of solid #NaHCO_3# produces 1 mole of #H_2O(g) #and 1 mole of #CO_2(g)# as volatile matter

Molar mass of
#NaHCO_3 =23+1+12+3*16=84g/"mol"#

Molar mass of
#CO_2=12+2*16=44g/"mol"#

Molar mass of #H_2O=2*1+16=18g/"mol"#

By the problem the decrease in mass = initial mass -residual mass= 100-90.7=9.3g

According to the decomposition reaction total 44+18=62g decrease in mass occurs for presence of #2*84g=168g# of #NaHCO_3# in the mixture.

So decrease in mass of 9.3 g will occur for the presence of

#(168*9.3)/62g=25.2g" "NaHCO_3#
in the given mixture

So the amount#Na_2CO_3=100-25.2=74.8g# in 100g mixture

So percentage of #Na_2CO_3 =74.8%#