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kramer
3 years ago
8

How many liters of carbon dioxide will 0.5 moles of lithium hydroxide (LiOH) absorb?

Chemistry
1 answer:
Elden [556K]3 years ago
5 0

Answer : The volume of carbon dioxide will be, 5.6122 L

Solution : Given,

Moles of LiOH = 0.5 moles

Molar mass of carbon dioxide = 44 g/mole

Density of carbon dioxide = 0.00196 g/ml

First we have to calculate the mass of carbon dioxide.

The balanced reaction will be,

2LiOH+CO_2\rightarrow Li_2CO_3+H_2O

From the reaction we conclude that

As, 2 moles of LiOH absorbs 44 grams of carbon dioxide

So, 0.5 moles of LiOH absorbs \frac{0.5moles}{2moles}\times 44g=11 grams of carbon dioxide

Mass of carbon dioxide = 11 g

Density of carbon dioxide = 0.00196 g/ml

Now we have to calculate the volume of carbon dioxide.

Density=\frac{Mass}{volume}

0.00196g/ml=\frac{11g}{volume}

Volume of carbon dioxide = 5612.24 ml = 5.6122 L      (1 L = 1000 ml)

Therefore, the volume of carbon dioxide will be, 5.6122 L

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Dafna11 [192]

Answer : The enthalpy for the reaction is 49.1 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_6H_6 will be,

6C(s)+3H_2(g)\rightarrow C_6H_6(l)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) 2C_6H_6(g)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)    

\Delta H_1=-6271kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    

\Delta H_2=-393.5kJ/mole

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    

\Delta H_3=-483.6kJ/mole

Now we will reverse the reaction 1 and divide by 2, multiply reaction 2 by 6 and reaction 3 by 3/2 then adding all the equations, we get :

(1) 6CO_2(g)+3H_2O(l)\rightarrow C_6H_6(g)+\frac{15}{2}O_2(g)    

\Delta H_1=-\frac{-6271kJ/mole}{2}=3135.5kJ/mol

(2) 6C(s)+6O_2(g)\rightarrow 6CO_2(g)    

\Delta H_2=6\times (-393.5kJ/mole)=-2361kJ/mol

(3) 3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)    

\Delta H_3=\farc{3}{2}\times (-483.6kJ/mole)=-725.4kJ/mol

The expression for enthalpy of formation of C_6H_6 will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(3135.5kJ/mole)+(-2361kJ/mole)+(-725.4kJ/mole)

\Delta H=49.1kJ/mole

Therefore, the enthalpy for the reaction is 49.1 kJ/mol

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The conclusion which can be drawn from the results that more ice added caused a greater change in temperature.

<h3>Why temperature of substance changes?</h3>

Temperature of any substance will change if some amount of energy is absorbed or released by that substance in the form of heat.

If in simple water, we add some amount of ice then energy or heat from the whole mixture will released as a result of which temperature of the water decreases. And decrease in temperature will depend on the amount of added ice.

Hence, more ice added caused a greater change in temperature.

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