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Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2:
2NaCl(s)+2H2SO4(l)+MnO2(s)→Na2SO4(s)+MnSO4(s)+2H2O(g)+Cl2(g)
Assume that the gas produced is saturated with water vapor at a partial pressure of 28.7 Hg and that it has a volume of 0.597L at 27 ∘C and 755 mm Hg pressure.

What is the mole fraction of Cl2 in the gas?

Respuesta :

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This problem is providing us with the chemical reaction used to prepare chlorine from sodium chloride, sulfuric acid and manganese (IV) oxide. Thus, the mole fraction of chlorine in the gaseous product is required, given the partial pressure of steam, the total volume, temperature and total pressure. The result is 0.962, according to the following:

Mole fraction:

In chemistry, relative compositions are used to give us an idea of the amount a substance has in a mixture. In the case of gases, we can use the pressure of each gas in order to calculate the mole fraction according to the Dalton's law:

[tex]x_i=\frac{P_i}{P} [/tex]

Thus, since we have the total pressure and the pressure of steam, we can calculate the partial pressure of chlorine as follows:

[tex]P_{Cl_2}=755mmHg-28.7mmHg=726.3mmHg[/tex]

Next, we can use the first equation to obtain the required:

[tex]x_i=\frac{726.3mmHg}{755mmHg} =0.962[/tex]

Learn more about Dalton's law: https://brainly.com/question/14119417

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