grams from a decomposition reaction?

Question

Determine how many grams of Na2CO3 (solid) (106 g/ mol) and CO2 (gas) (44g/mol) are produced from the decomposition of 1.67 grams of NaHCO3 (solid) (84 g/ mol).


thanks in advance!!!

Answer 1

2NaHCO3(s) --> Na2CO3(s) + CO2(g) + H2O(l)

Na2CO3 MW = 105.99 g/mole
NaHCO3 MW = 84.007 g/mole
CO2 MW = 44.009 g/mole

1.67 g NaHCO3 = 1.67 g * mole/84.007g = 0.01988 moles NaHCO3

2NaHCO3 --> 1 Mole product

so 0.01988/2 = 0.009939 mole products expected

Na2CO3

0.009939 mole Na2CO3 = 0.009939 mole * 105.99 g/mole =

= 1.05 g Na2CO3

CO2

0.009939 mole CO2 = 0.009939 mole * 44.009 g/mole =

= 0.44 g CO2

Answer 2

2NaHCO3 ===> Na2CO3 + CO2 + H2O

1.67gNaHCO3 x 1molNaHCO3/84gNaHCO3 x 1molNa2CO3/2molNaHCO3 x 106gNa2CO3/1molNa2CO3 = 1.05g Na2CO3

1.67gNaHCO3 x 1molNaHCO3/84gNaHCO3 x 1molCO2/2molNaHCO3 x 44gCO2/1molCO2 = 0.44g CO2