The carbon monoxide present in a gas sample can react with solid diiodine pentox

Question

The carbon monoxide present in a gas sample can react with solid diiodine pentoxide

to produce gaseous carbon dioxide and gaseous diatomic iodine. The iodine can then

be bubbled through excess aqueous sodium thiosulfate to form aqueous iodide and

aqueous tetrathionate ions, S4O6

2?

. By back titrating the excess sodium thiosulfate, the

concentration of carbon monoxide in the original sample can be determined.

An impure 18.7 mg gas sample was passed over solid diiodine pentoxide. The carbon

dioxide produced was then bubble through 10.32 mL of 0.02459 M sodium thiosulfate

solution. Finally, the excess sodium thiosulfate was back titrated with 4.480 mL of

4.37x10?3

M I2 solution. Write the balanced chemical equation for each reaction and

determine the weight percent of carbon monoxide in the original gas sample.

Explanation / Answer

The balanced equations are:

5 CO(g) + I2O5(s) = 5 CO2(g) + I2(g)

I2(g) + 2 Na2S2O3(aq) => Na2S4O6(aq) + 2 NaI(aq)


Total moles of Na2S2O3 = volume x concentration of Na2S2O3

= 10.32/1000 x 0.02459 = 0.0002537688 mol


Moles of excess Na2S2O3 = 2 x moles of I2 from I2 solution

= 2 x volume x concentration of I2 solution

= 2 x 4.480/1000 x 4.37 x 10^(-3) = 3.91552 x 10^(-5) mol


Moles of reacted Na2S2O3 = total - excess moles of Na2S2O3

= 0.0002537688 - 3.91552 x 10^(-5) = 0.0002146136 mol


Moles of I2 produced in reaction = 1/2 x moles of reacted Na2S2O3

= 1/2 x 0.0002146136 = 0.0001073068 mol


Moles of CO = 5 x moles of I2 produced

= 5 x 0.0001073068 = 0.000536534 mol


Mass of CO = moles x molar mass of CO

= 0.000536534 x 28.01 = 0.015028 g = 15.028 mg


Weight% of CO = mass of CO/mass of sample x 100%

= 15.028/18.7 x 100%

= 80.37% = 80.4%

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