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anyanavicka [17]
2 years ago
5

Molecular chlorine and molecular fluorine combine to form a gaseous product. Under the same conditions of temperature and pressu

re it is found that one volume of cl2 reacts with three volumes of f2 to yield two volumes of the product. What is the formula of the product?
Chemistry
1 answer:
uranmaximum [27]2 years ago
8 0

Answer:

  • Cl F₃

Explanation:

<u>1) Reactants:</u>

The reactants are:

  • <em>Molecular chlorine</em>: this is a gas diatomic molecule, i.e. Cl₂ (g)

  • <em>Molecular fluorine</em>: this is also a gas diatomic molecule: F₂ (g)

<u>2) Stoichiometric coefficients:</u>

  • <em>One volume of Cl₂ react with three volumes of F₂</em> means that the reaction is represented with coefficients 1 for Cl₂ and 3 for F₂. So, the reactant side of the chemical equation is:

        Cl₂ (g) + 3F₂ (g) →

<u>3) Product:</u>

  • It is said that the reaction yields <em>two volumes of a gaseous product;</em> then, a mass balance indicates that the two volumes must contain 2 parts of Cl and 6 parts of F. So, one volume must contain 1 part of Cl and 3 parts of F. That is easy to see in the complete chemical equation:

       Cl₂ (g) + 3F₂ (g) → 2Cl F₃ (g)

        As you see, that last equation si balanced: 2 atoms of Cl and 6 atoms of F on each side, and you conclude that the formula of the product is ClF₃.

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A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
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Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

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n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

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T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

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P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

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