Question #bfd77

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

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%`