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earnstyle [38]
3 years ago
6

How many moles of NaCl are produced if 239.7 grams of Na2S reacts with plenty of AlCl3?

Chemistry
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

6.142 moles of NaCl

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AlCl3 + 3Na2S —> Al2S3 + 6NaCl

Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:

Mass mass of Na2S = 78.048g/mol

Mass of Na2S = 239.7g

Number of mole Na2S =..?

Mole = Mass /Molar Mass

Number of mole Na2S = 239.7/78.048 = 3.071 moles

Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:

From the balanced equation above,

3 moles of Na2S reacted to produce 6 moles of NaCl.

Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl

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He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
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Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

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\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

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Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

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Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

8 0
3 years ago
Read 2 more answers
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