A 10.00 gram mixture of NaHCO3 and Na2CO3 was heated and yielded 0.0357 moles of H2O...

A 10.00 gram mixture of NaHCO3 and Na2CO3 was heated and yielded 0.0357 moles of H2O and 0.1091 moles of CO2. The reactions are described by: NaHCO3 —> Na2O +CO2+H2O and Na2CO3 —> Na2O + CO2. Calculate the volume of CO2 produced in EACH reaction at 733 Torr and 43 degrees C and the percentage composition of the mixture.

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Solutions

Expert Solution

the balanced reaction for both are:

2NaHCO3 ------------> Na2O + 2 CO2 + H2O

Na2CO3 ---------------> Na2O + CO2

From two reactions we can clearly observe that only in reaction-1 water is produced hence 0.0357 moles of water is produced from reaction-1. Observing their co-efficients, when 2 moles of CO2 are produced 1 mole of H2O is produced.

Therefore when 0.0357 moles of H2O is produced, 2*0.0357 = 0.0714 moles of CO2 is produced.

also amount of NaHCO3 present = 2*0.0357 = 0.0714 moles=0.0714*84 =5.99 grams

Remaining CO2 is produced from other equation.

Amount of CO2 released from Na2CO3 = 0.1091 - 0.0714 =0.0377 moles

amount of Na2CO3 = 0.0357 moles = 0.0377*105.9 =3.99 grams

ratio of NaHCO3 to Na2CO3 = 5.99/3.99 = 1.5

pressure=743 torr = 743/760 atm =0.977 atm

temperature = 43 degrees = 43+273 = 316 K

Ideal gas equation PV =nRT

volume of CO2 released by NaHCO3 V = nRT/P = 0.0714*0.0821*316/0.977 = 1.895 lit

volume of CO2 released by Na2CO3 V = nRT/P = 0.0377*0.0821*316/0.977 = 1.001 lit

queen_honey_blossom answered 1 year ago

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