Answer to Question #154828 in General Chemistry for Ashley

Question

Is my answer right for this question?

1. Assuming the conditions are at SATP, calculate both the theoretical volume and mass (in g) of both water and carbon dioxide produced (assuming all of the sodium bicarbonate decomposed).  Use proper sig digs. Show all of your work.  (/4)

2 NaHCO3  →   Na2CO3  + H2O  +   CO2

3.95g Na2CO3 x 1 mol Na2CO3 / 84.01g Na2CO3 = 0.0470 mol Na2CO3

H2O

0.0470 mol Na2CO3 x 1 mol H2O / 1 mol Na2CO3 = 0.0470 mol H2O

0.0470 mol H2O x 18.02 g H2O / 1 mol H2O = 0.847g H2O (MASS, 3SD)

                               (1.01 x 2 + 16.00)

0.847g H2O x 1ml H2O / 1g H2O = 0.847ml H2O (VOLUME, 4SD)

CO2

0.0470 mol Na2CO3 x 1 mol CO2 / 1 mol Na2CO3 = 0.0470 mol CO2 (n, gas law)

0.0470 mol CO2 x 44.01g CO2 / 1 mol CO2 = 2.07g CO2 (MASS, 3SD)

                               (16.00 x 2 + 12.01)

PV = nRT 

V = nRT / P 

V = (0.0470 mol CO2) (8.314) (24.8) / 101.325 

V = 0.0956L CO2 (VOLUME, 3SD) 

Answer 1

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