19 g of a mixture containing textNaHC textO text3 and Na2CO3 on complete heating liberated 1.12 L of CO2 at STP. The weight of the remaining solid was 15.9 g. What is the weight (in g) of Na2CO3 in the mixture before heating?

Question

19 g of a mixture containing textNaHC textO text3 and Na2CO3 on complete heating liberated 1.12 L of CO2 at STP. The weight of the remaining solid was 15.9 g. What is the weight (in g) of Na2CO3 in the mixture before heating? (A) 8.4 (B) 15.9 (C) 4.0 (D) 1.6

Tardigrade Admin · Accepted Answer

Molecular weight of NaHCO3=23+1+12+48=84 Molecular weight of Na2CO3=46+12+48=106 Hence, total weight =84+106=190 ∵ In 190 g of a mixture, weight of NaCO3 is =106 ∴ In 19 g of a mixture weight of Na2CO3=(106× 19/190) =10.6g

Q. 19 g of a mixture containing NaHCO3​ and Na2​CO3​ on complete heating liberated 1.12 L of CO2​ at STP. The weight of the remaining solid was 15.9 g. What is the weight (in g) of Na2​CO3​ in the mixture before heating?

8.4

15.9

4.0

1.6

Molecular weight of NaHCO3​=23+1+12+48=84 Molecular weight of Na2​CO3​=46+12+48=106 Hence, total weight =84+106=190 ∵ In 190 g of a mixture, weight of NaCO3​ is =106 ∴ In 19 g of a mixture weight of Na2​CO3​=190106×19​ =10.6g

Q. 19 g of a mixture containing $NaHC O_{3}$ and $N a_{2} C O_{3}$ on complete heating liberated 1.12 L of $C O_{2}$ at STP. The weight of the remaining solid was 15.9 g. What is the weight (in g) of $N a_{2} C O_{3}$ in the mixture before heating?