Question

According to the following reaction, how much energy is evolved during the reaction of 32.5 g B2H6 and 72.5 g Cl2? The molar mass of B2H6 is 27.67 g/mol. B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g) ΔH°rxn = -1396 kJ According to the following reaction, how much energy is evolved during the reaction of 32.5 g B2H6 and 72.5 g Cl2? The molar mass of B2H6 is 27.67 g/mol. B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g) ΔH°rxn = -1396 kJ a) 1640 kJ b) 1430 kJ c) 429 kJ d) 3070 kJ e) 238 kJ.

Answer 1

Answer: The amount of heat evolved during the reaction is 238 kJ

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For $B_2H_6$ :

Given mass of $B_2H_6$ = 32.5 g

Molar mass of $B_2H_6$ = 27.67 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_2H_6=\frac{32.5g}{27.67g/mol}=1.174mol$

• For chlorine gas:

Given mass of chlorine gas = 72.5 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{72.5g}{71g/mol}=1.021mol$

For the given chemical equation:

$B_2H_6(g)+6Cl_2(g)\rightarrow 2BCl_3(g)+6HCl(g)$

By Stoichiometry of the reaction:

6 moles of chlorine gas reacts with 1 mole of $B_2H_6$

So, 1.021 moles of chlorine gas will react with = $\frac{1}[6}\times 1.021=0.170mol$ of $B_2H_6$

As, given amount of $B_2H_6$ is more than the required amount. So, it is considered as an excess reagent.

Thus, chlorine gas is considered as a limiting reagent because it limits the formation of product.

Moles of $B_2H_6$ reacted = 0.170 moles

Amount of heat released for 1 mole of $B_2H_6$ = 1396 kJ

So, amount of heat released for 0.170 moles of $B_2H_6=\frac{1396}{1}\times 0.170=238kJ$

Hence, the amount of heat evolved during the reaction is 238 kJ

Answer 2

The amount of energy that is evolved during the reaction of 32.5 grams of $B_2H_6$ and 72.5 grams of $Cl_2$ is: e) 238 kJ.

Given the following data:

• Mass of $B_2H_6$ = 32.5 grams

• Mass of $Cl_2$ = 72.5 grams

• Molar mass of $B_2H_6$ = 27.67 g/mol.

• Enthalpy of combustion = -1396 kJ/mol

To find the amount of energy that is evolved during the reaction of 32.5 grams of $B_2H_6$ and 72.5 grams of $Cl_2$:

First of all, we would determine the number of moles of $B_2H_6$ in this chemical reaction.

                             $B_2H_6_{(g)} + 6Cl_2_{(g)}$ ----> $2BCl_3_{(g)} + 6HCl_{(g)}$

$Number\;of\;moles \;(B_2H_6)= \frac{Mass\; of\;B_2H_6}{Molar\;mass\;of\;B_2H_6}$

$Number\;of\;moles \;(B_2H_6)= \frac{32.5}{27.67}$

Number of moles ($B_2H_6$) = 1.1746 moles.

For $Cl_2$:

$Number\;of\;moles \;(Cl_2)= \frac{Mass\; of\;Cl_2}{Molar\;mass\;of\;Cl_2}$

$Number\;of\;moles \;(Cl_2)= \frac{72.5}{71}$

Number of moles ($Cl_2$) = 1.0211 moles.

6 moles of $Cl_2$ = 1 mole of $B_2H_6$

1.0211 mole of $Cl_2$ = X mole of $B_2H_6$

Cross-multiplying, we have:

$6X = 1.0211\\\\X = \frac{1.0211}{6}$

X = 0.1702 moles of $B_2H_6$

Now, we can find the amount of energy that is evolved during the reaction of 32.5 grams of $B_2H_6$ and 72.5 grams of $Cl_2$:

1 mole of $B_2H_6$ = 1396 kJ/mol

0.1702 mole of $B_2H_6$ = Y kJ/mol

Cross-multiplying, we have:

$Y = 0.1702(1396)$

Y = 237.6 ≈ 238 kJ.

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