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liq [111]
3 years ago
13

7. Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. What is the yield of S2Cl2 expected in a laborat

ory experiment from the reaction of 10.0 g S8 and 6.00 g Cl2? What if the reaction consistently gives an 85.0 % yield?
Chemistry
1 answer:
Pie3 years ago
3 0

Answer:

11.4g of S₂Cl₂ is the expected yield

9.69g of S₂Cl₂ are produced with a 85% yield

Explanation:

The reaction of sulfur S₈ with Cl₂ to produce S₂Cl₂ is:

S₈ + 4Cl₂ → 4S₂Cl₂

<em>Where 1 mole of sulfur reacts with four moles of chlorine to produce four moles of disulfur dichloride.</em>

To find the limiting reactant you need to convert mass of each reactant to moles using its molar mass, thus:

S₈ (Molar mass: 256.52g/mol): 10.0g ₓ (1mol / 256.52g) = 0.0390 moles S₈

Cl₂ (Molar mass: 70.9g/mol): 6.00g ₓ (1mol / 70.9g) = 0.0846 moles Cl₂

For a complete reaction of 0.0390 moles of sulfur, there are necessaries:

0.0390 mol S₈ ₓ (4 mol Cl₂ / 1 mol S₈) = <em>0.156 moles Cl₂. </em>As you have just 0.0846 moles of chlorine, Cl₂ is the limiting reactant.

As 4 moles of Cl₂ produce 4 moles of S₂Cl₂.<em> 0.0846 moles of Cl₂ produce, in theory, 0.0846 moles of S₂Cl₂ (Molar mass: 135.04g/mol). </em>In mass:

0.0846 moles S₂Cl₂ ₓ (135.04g/mol) =

<h3>11.4g of S₂Cl₂ is the expected yield</h3>

If you produce just the 85.0% of yield, mass of S₂Cl₂ is:

11.4g ₓ 85% =

<h3>9.69g of S₂Cl₂</h3>
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How to prepare magnesium carbonate starting with magnesium nitrate
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How to prepare Magnesium Carbonate:

Explanation:

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8 0
3 years ago
For the reaction CO+2H2=CH3OH at 700 K, equilibrium concentrations are [H2]=0.072 M, [CO]= 0.020M, and [CH3OH]= 0.030 M. Calcula
bekas [8.4K]

The balanced equation for the reaction is

CO(g) + 2H₂(g) ⇄ CH₃OH(g)

 

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

   k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²

</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M

 

From substitution,

   k = 0.030 M / 0.020 M x (0.072 M)²

   k = 289.35 M⁻²

<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².

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3 0
3 years ago
50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the
scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

H_2NC_2H_5CO^-_2  +  H^+  ------>  H_3}^+NC_2H_5CO^-_2

1 mole of  alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL  of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via  the following process as shown below;

[H_3}^+NC_2H_5CO^-_2] = \frac{initial moles of alaninate}{total volume}

[H_3}^+NC_2H_5CO^-_2] = \frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}

[H_3}^+NC_2H_5CO^-_2] = \frac{8}{100mL}

[H_3}^+NC_2H_5CO^-_2] = 0.08 M

Alanine serves as an intermediary form, however the concentration of H^+ and the pH can be determined as follows;

[H^+] = \sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{  K_{a1}{[H_3}^+NC_2H_5CO^-_2]  } }

[H^+] = \sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})}  {(10^{-pK_{a1}})+(0.08)} }

[H^+] = \sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})}  {(10^{-2.344})+(0.08)} }

[H^+] =  7.63*10^{-7}M

pH = - log [H^+]

pH = -log[7.63*10^{-7}]

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

H_2NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO^-_2

H^+_3NC_2H_5CO^-_2    +  H^+     ----->   H^+_3NC_2H_5CO_2H

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

[H^+_3NC_2H_5CO_2H] = \frac{initial moles of alaninate}{total volume}

[H^+_3NC_2H_5CO_2H] = \frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}

[H^+_3NC_2H_5CO_2H] = \frac{8}{150}

[H^+_3NC_2H_5CO_2H] = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

H^+_3NC_2H_5CO_2H        ⇄        H^+_3NC_2H_5CO^-_2    +  H^+

(0.053 - x)                                  x                             x

K_{a1} = \frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}

10^{-PK_{a1}} = \frac{x*x}{(0.053-x)}

10^{-2.344} =\frac{x^2}{(0.053-x)}

0.00453 = \frac{x^2}{(0.053-x)}

0.00453(0.053-x) =x^2

x^2+0.00453x-(2.4009*10^{-4})

Using quadratic equation formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

we have:

\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)} OR \frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}

= 0.0134                    OR                -0.0179

So; we go by the positive integer which says

x = 0.0134

So [H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M

pH = -log[H^+]

pH = -log[0.0134]

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0
3 years ago
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