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Serjik [45]
3 years ago
9

_S8+ _02 - SO3

Chemistry
1 answer:
Anton [14]3 years ago
6 0

Answer:

  • a)       S_8+12O_2\rightarrow 8SO_3
  • b)       <em>Combination or synthesis reaction</em>
  • c)

               1molS_8:12molO_2;or\\ \\ \\ \dfrac{1molS_8}{12molO_2}

  • d) <em>30 g.</em>

Explanation:

<em><u>a. Balance the Equation</u></em>

To balance the equation you must add the coefficients that yield the same number of atoms for the same kind of atoms on each side of the chemical equation.

You cannot modify the superscripts because the they are part of the chemical formula of the compound or molecule, they indicate the number of atoms of each kind inside the compound, i.e. they determine the composition substance.

The coefficients can be adjusted because they just indicate the number of units that react or are produced.

Starting from the skeleton equation:

         S_8+O_2\rightarrow SO_3

Add an 8 before the SO₃ to balance the S atoms:

       S_8+O_2\rightarrow 8SO_3

Now add a coefficient of 12 for the O₂ molecule on the left side:

       S_8+12O_2\rightarrow 8SO_3

If you count:

Atom     Leff side      Right side

   S            8                      8

   O           24                    24

Thus, the equation is balanced.

<em><u>b. What type of chemical reaction is this?</u></em>

When two elements or compounds combine to form one single compound that is a combination reaction.

The general form of a combination reaction is:

  • A + B → AB

This is called comibnation to distinguish from single replacement, double replacement, decompositon, and combustion reaction.

It is also named synthesis reaction because a new compound is being synthetized ("created") from simpler substances.

The combination of a non-metal and oxygen is a typical reaction of synthesis in which a basic oxide is produced. They are named basic oxides because they react with water to form a base: hydroxide.

<em><u>c. What is the molar ratio of S₈, to O₂?</u></em>

<u />

The mole ratio also called theoretical or stoichiometric mole ratio is the ratio of coefficients of the substances. The mole ratio of the reactants indicate the proportion in which the reactants combine to each other to form the product or products.

Then, take the coefficients of S₈ and O₂ from the balanced molecular chemical equation and form the ratio:

            1molS_8:12molO_2;or\\ \\ \\ \dfrac{1molS_8}{12molO_2}

That means that every molecule (or mole of molecules) of S₈ reacts with 12 molecules (or mole of molecules) of O₂ to form the products: in this case 8 molecules (or mole of molecules of SO₃).

You use this ratio to determine the yield of the reactions in a procedure called stoichiometric calculations.

<em><u>d. How many grams of O₂ should be used if 100 grams of SO₃, is the desired end product?</u></em>

You will need the mole ratio of O₂ to SO₃ and the molar masses of both substanes.

Mole ratio:

           \dfrac{8molSO_3}{12molO_2}

Moles of SO₃

  • moles = mass in grams/molar mass
  • molar mass of SO₃ = 80.066g/mol
  • moles of SO₃ = 100g / 80.066g/mol = 1.25 mol SO₃

Moles of O₂

        1.25molSO_3\times 12molO_2/8molSO_3=1.875molO_2

Conver to mass of O₂

  • molar mass of O₂ = 15.999 g/mol
  • mass of O₂ = 1.875mol × 15.999g/mol = 29.998 g
  • Round to 1 significant figures: 30 g.

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A solution was prepared by dissolving 195.0 g of KCl in 215 g of water. Calculate the mole fraction of KCl. (The formula weight
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Answer:

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Explanation:

The mole fraction of a compound in a solution is:

\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}.

In this question, the mole fraction of \rm KCl in this solution would be:

\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}.

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\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})}\end{aligned}.

From the question:

  • Mass of \rm KCl: m(\mathrm{KCl}) = 195.0\; \rm g.
  • Molar mass of \rm KCl: M(\mathrm{KCl}) = 74.6\; \rm g \cdot mol^{-1}.
  • Mass of \rm H_2O: m(\mathrm{H_2O}) = 215\; \rm g.
  • Molar mass of \rm H_2O: M(\mathrm{H_2O}) = 18.0\; \rm g\cdot mol^{-1}.

Apply the formula \displaystyle n = \frac{m}{M} to find the number of moles of \rm KCl and \rm H_2O in this solution.

\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= \frac{195.0\; \rm g}{74.6\; \rm g \cdot mol^{-1}} \approx 2.61\; \em \rm mol\end{aligned}.

\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= \frac{215\; \rm g}{18.0\; \rm g \cdot mol^{-1}} \approx 11.9\; \em \rm mol\end{aligned}.

The molar fraction of \rm KCl in this solution would be:

\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})} \\ &\approx \frac{2.61 \; \rm mol}{2.61\; \rm mol + 11.9\; \rm mol} \approx 0.180\end{aligned}.

(Rounded to three significant figures.)

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