Question

NAME POSTLABORATORY ASSIGNMENT SECTION 8 sample of potassium hydrocarbonate is decomposed by heating potassium carbonate Caksulate the theoretical vield and percent yield of onate is decomposed by heating to produce 2 KHCO3(-) K 00,6) + H2O) + coce) 2 A 1175 sample mixture of potassium hydropen carbonate is decomposed by heating of the malo is 0.275 g. what is the percentage of KHCO, in the unknown mixture 2 KHCO - KC0,0) + HCO3(e)

Answer 1

theoretical yield =

$1.225\ g\ KHCO_3\times \frac{1\ mol\ KHCO_3}{100.115g}\times \frac{1\ mol\ K_2CO_3}{2\ mol\ KHCO_3}\times \frac{138.2\ g}{1\ mol\ K_2CO_3}$

=0.845 g

Experimental yield = 0.855 g

Percentage yield =

$=\frac{experimental\ yield}{theoretical\ yield}\times 100$

$=\frac{0.855}{0.845}\times 100$

= 101.2%

--------------------------------------------------------------------

2KHCO3(s) <--> K2CO3(s) + H2CO3(g)

Mass lost corresponds to H2CO3, becase it is the only gaseous product

Mass of H2CO3 = 0.275 g

Mass of KHCO3 present in mixture =

$0.275\ g\ H_2CO_3\times \frac{1\ mol\ H_2CO_3}{62\ g}\times \frac{2\ mol\ KHCO_3}{1\ mol\ H_2CO_3}\times \frac{100.115\ g}{1\ mol\ KHCO_3}$

=0.888 g

Percentage of KHCO3 in the mixture =

$=\frac{mass\ of \ the\ compound}{total\ mass\ of\ mixture}\times 100$

$=\frac{0.888\ g}{1.775\ g}\times 100$

= 50.02%