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Crazy boy [7]
3 years ago
9

Calorimetric studies show that the reaction is exothermic. 2NO2(g) N2O4(g) + 14.1 kcal. Based on this information, which one--if

any--of the following additional changes would increase the molar concentration at equilibrium of N2O4(g)?
decrease the pressure

increase the temperature

decrease the concentration of NO2(g)

stir the reaction mixture

none of the above
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
5 0
In order for you to get the answer, please have in mind the following situation: To increase the molar concentration of N2O4(g), 2NO2(g) should also increase for equilibrium to occur. Now, this equation is exothermic. By <span>Le Chatelier's principle, equilibrium constant and reaction constants also come into play in terms of increasing or decreasing the temperature. After that I know you can find the answer. </span>
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If you need to reverse the following reaction and multiply it by 2 in order for it to be an intermediate reaction in a Hess's la
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The reaction is:
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mrs_skeptik [129]

Answer: The correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

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Moles=\frac{\text{Given mass}}{\text{Molar mass}}

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Moles=\frac{0.40g}{134.45g/mol}=0.00297mol

For the given chemical equation:  

4Al(s)+3CuCl_2(aq.)\rightarrow 2Al_2Cl_3(aq.)++3Cu(s)

By Stoichiometry,

3 moles of Copper chloride reacts with 4 moles of aluminium

So, 0.00297 moles of copper chloride will react with = \frac{4}{3}\times 0.00297 = 0.00396 moles of Aluminium.

As, the required moles of aluminium is less than the given moles of aluminium. Hence, it is considered as the excess reagent.

Copper chloride is the limiting reagent in the given chemical reaction because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of copper chloride produces 3 moles of copper metal.

So, 0.00297 moles of copper chloride will produce = \frac{3}{3}\times 0.00297 = 0.00297 moles of copper metal.

Now, to calculate the given mass of copper metal, we use the equation required to calculate moles:

Molar mass of copper = 63.5 g/mol

Putting values in that equation, we get:

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Given mas of copper metal = 0.188grams (approx 0.20grams)

For the given reaction, some of the aluminum is left behind because it is the excess reagent in the reaction.

Hence, the correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

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