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3.5.1a Vitamin C has the formula CxHyOz. You burn 0.589 g of the compound in a...

3.5.1a

Vitamin C has the formula CxHyOz. You burn 0.589 g of the compound in a combustion analysis chamber and isolate 0.883 g of CO2 and 0.240 g of H2O.

What is the amount of CO2 produced in the analysis, in moles?

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Answer #1

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass(CO2)= 0.883 g

use:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(0.883 g)/(44.01 g/mol)

= 2.006*10^-2 mol

Answer: 2.01*10^-2 mol

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