NaCl is stable in the temperature range between 200 o C and 850 o C, but in that

Question

NaCl is stable in the temperature range between 200 oC and 850 oC, but in that range, NaHCO3decomposes by the reaction:

2 NaHCO3 (s) Na2CO3 (s) + H2O (g) + CO2 (g)

If a mixture of the two is heated to temperatures in the above range, the gaseous products will escape and the residue will contain the original NaCl and an amount of solid Na2CO3. The Na2CO3 is related stoichiometrically to the amount of NaHCO3 which has decomposed. If the reaction has proceeded to completion (no remaining NaHCO3), the moles, and therefore the weight of NaHCO3originally in the sample, are able to be determined.

We heat the sample several times, weighing the residue after each heating.

We use the following criterion to establish that the reaction is complete:

The reaction is considered complete if the change in weight of the residue

is less that 5.0 mg (plus or minus) in two successive heatings.

The following data were collected in such a determination:

What was the weight in grams of the sample before the first heating?

What was the weight in grams of NaHCO3 in the original sample?

Explanation / Answer

Given chemical equation,

2NaHCO3 ----> Na2CO3 + H2O + CO2

weight of sample before the first heating = 15.1735 - 13.6888 = 1.4847 g

Final weight of crucible + residue = 14.8171 g

So, final weight of residue would be = 14.8171 - 13.6888 = 1.1283 g of Na2CO3

weight lost as (H2O + CO2)g = (1.4847) - (14.8171-13.6888) = 0.3564 g

H2O : CO2 = 18.015 g : 44.01 g = 0.29 : 0.71

So, mass of H2O lost = 0.29 x 0.3564 = 0.1034 g

mass of CO2 lost = 0.71 x 0.3564 = 0.2530 g

moles of H2O = /18.015 = 0.0103 mole

moles of NaHCO3 = 2 x 0.0103 = 0.02065 mole

mass of NaHCO3 = moles x molar mass = 0.02065 x 84.007 = 1.735 g

thus, mass of NaHCO3 in the original sample is 1.735 g

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.