Answer to Question #130026 in Organic Chemistry for Priyanshu

Question

Heating 2g of a mixture of na2co3 and nahco3 at 300c reduces the load to 0.248. What is the percentage amount of na2co3 in the mixture?

Answer 1

$Na_2CO_3(s)$ does not decompose$2NaHCO_3(s) \rightarrow Na_2CO_3 (s)+ CO_2(g) + H_2O(g)$

The reduce in load is due to release of CO2 and H2O

Let the moles of CO2 to be unnkown, to be x. Convert moles of CO2 to mass of CO2

$x moles (CO_2)\times \frac{44.01 g(CO_2)}{1 mole (CO_2)}= x\times 44.01 grams (CO_2)$

Convert moles of CO2 to moles of H2O

$x moles (CO_2)\times \frac{1 mole (H_2O)}{1 mole (CO_2)}= x moles (H_2O)$

Convert moles of H2O to mass of H2O

$x moles (H_2O)\times \frac{18.02 g (H_2O)}{1 mole (H_2O)}= x\times 18.02 grams (H_2O)$

The sum of the masses of CO2 and H2O is equal to the mass loss, i.e. to 0.248 grams

$m(H_2O)+ m(CO_2) =0.248$

$18.02\times x + 44.01\times x =0.248$

$x=$0.003998 mol

So, moles of CO2 = 0.003998 moles

Convert moles of CO2 to moles of NaHCO3

$0.003998 moles (CO_2) \times \frac{2 moles (NaHCO_3)}{1 mole (CO_2)}=0.007996 moles (NaHCO_3)$

Convert moles of NaHCO3 to mass of NaHCO3

$0.007996 moles (NaHCO_3)\times \frac{84.01 g (NaHCO_3)}{1 mole (NaHCO_3)}= 0.672 g (NaHCO_3)$

Find mass of Na2CO3

$m(Na_2CO_3)= m_{total}-m(NaHCO_3)= 2-0.672 = 1.33 g (Na_2CO_3)$

mass percent Na2CO3 =$\frac{m(Na_2CO_3)}{m_{total}}\times 100\% = \frac{1.33 g}{2 g}\times 100\% = 66.4\%$

Answer: 66.4 %