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anyanavicka [17]
3 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]3 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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Answer:

-21 kJ·mol⁻¹  

Explanation:

Data:

                    H₃O⁺ +  OH⁻ ⟶ 2H₂O

       V/mL:    50         50  

c/mol·dm⁻³:   1.0         1.0

     

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\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

     nΔH   +         mCΔT       + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 +   50×4.5  = 0

0.050ΔH +          1883        +      225    = 0

                                  0.050ΔH + 2108 = 0

                                              0.050ΔH = -2108

                                                        ΔH = -2108/0.0500

                                                              = -42 000 J/mol

                                                              = -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

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