The student conducts an experiment to determine the composition of a mixture of

Question

The student conducts an experiment to determine the composition of a mixture of NaHCO3 (Molar Mass 84.01 g/mol) and Na2CO3 (MM 105.99) The student places a sample of the mixture into a preweighed test tube that is attached to a container that holds a drying agent. The student heats the test tube with a bunson burner during which the water produced by the rxn is captured by the drying agent.

Mass of empty tube = 15.825g

Mass of test tube and mixture before heating= 17.648g

Mass of drying agent before rxn= 2.134 g

Mass of drying agent and water= 2.303g

a. Calculate the number of moles of NaHCO3(s) present in the mixture before rxn was initiated.

b. Determine mass percent of NaHCO3(s) in the mixture

c. if the student spills some of the mixture out the test tube after weighing the test tube and mixture before heating, how would this affect the mass percent of NaHCO3 calculated in part b. Justify answer

Determine the mass percent of NaHcols) in the mixture. d. 1 pointl if the student spils some of the mixture out of the test tube after weighing the test tube and mixture and before heating how would this e affect the mass percent of NaHco, calculated in part lch? Justify Your answer. When a sample of pure Nazco is placed in distilled water, the student observes that the pH of the solution increases significantly. This process is reprsented by the balanced net-ionic equal OH laq the Na,coalagl solution. (1 point write the esxpression for Kb for the carboante ion. E. pointl Calcualte the value of Rt for the carbonate ion.

Explanation / Answer

When a mixture of NaHCO3 and Na2CO3 are heated only NaHCO3 will decompose to give H2O and CO2 whereas Na2CO3 will remain as it is.

So no. of moles of water produced = no. of moles of NaHCO3 decomposed

given:

Mass of empty tube = 15.825g

Mass of test tube and mixture before heating= 17.648g

Mass of NaHCO3 and Na2CO3 = 17.648 - 15.825 = 1.823 g

Mass of drying agent before heating = 2.134 g

Mass of drying agent after heating = 2.303 g

Mass of water produced by decomposition = 2.303 g - 2.134 g = 0.169 g

Molar mass of water = 18 g/mol

No. of moles of water = Mass of water / molar mass of water = 0.169 g / 18 g/mol = 0.009389 moles

So no. of moles of NaHCO3 decomposed = 0.009389 moles

Molar mass of NaHCO3 = 84 g/mol

mass of NaHCO3 decomposed = No. of moles of NaHCO3 decomposed * Molar mass of NaHCO3 decomposed

= 0.009389 moles * 84 g/mol = 0.788 g of NaHCO3 has decomposed

(a)

Mass of NaHCO3 decomposed = 0.788 g Answer

(b) mass percent of NaHCO3(s) in the mixture = Mass of NaHCO3 / Mass of NaHCO3 and Na2CO3

= 0.788 g /1.823 g = 0.432

mass percent of NaHCO3(s) in the mixture = 43.22 % Answer

(c) if he spills out the mixture after weighing and before heating mass of mixture will remain the same but mass of NaHCO3(s) will be reduced hence it wil reduce the mass percent of NaHCO3(s) in part (b)

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