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Chapter 6: Problem 119

Consider the reaction \(\mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-2035 \mathrm{~kJ}\) Calculate the amount of heat released when \(54.0 \mathrm{~g}\) of diborane is combusted.

Short Answer

Expert verified
The amount of heat released when 54.0 g of diborane is combusted is -4070 kJ.

Step by step solution

01

Calculate the number of moles of diborane

We have 54.0g of diborane (B2H6), and we need to find the number of moles. Use the molar mass of diborane to convert mass to moles: Molar mass of B2H6 = 2 × (Molar mass of Boron) + 6 × (Molar mass of Hydrogen) Molar mass of B2H6 = 2 × (10.81 g/mol) + 6 × (1.01 g/mol) Moles of B2H6 = \( \frac{mass}{molar~mass} \) Moles of B2H6 = \( \frac{54.0 g}{2 \times 10.81 ~ g/mol + 6 \times 1.01 ~ g/mol} \)
02

Calculate the heat released during the combustion

The balanced chemical equation states that when 1 mole of B2H6 is reacted, the heat released is -2035 kJ. Now we need to find the heat released for the number of moles of B2H6 that we calculated in step 1. Heat released = moles of B2H6 × heat released per mole Heat released = (moles of B2H6) × (-2035 kJ) Note that the negative sign in the heat value indicates that heat is released (exothermic reaction).
03

Calculate the final answer

Now we have all the values needed to find the amount of heat released during the combustion. Perform the calculations using the values obtained in Step 1 and Step 2: Moles of B2H6 = \( \frac{54.0 g}{2 \times 10.81 ~ g/mol + 6 \times 1.01 ~ g/mol} \) = 2 moles Heat released = 2 moles × (-2035 kJ) = -4070 kJ Hence, the amount of heat released when 54.0 g of diborane is combusted is -4070 kJ.

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Most popular questions from this chapter

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5 \mathrm{~L}\). Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.

In a coffee-cup calorimeter, \(150.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) is added to \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) to make \(200.0 \mathrm{~g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\). If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{~kJ} / \mathrm{mol}\), calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is \(4.184 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and assume no heat loss to the surroundings.

It has been determined that the body can generate \(5500 \mathrm{~kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise? (The heat of vaporization of water is \(40.6 \mathrm{~kJ} / \mathrm{mol}\).)

The bomb calorimeter in Exercise 106 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A 1.056-g sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right.\) ) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{~kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is \(-803 \mathrm{~kJ} / \mathrm{mol}\). Use these data and Hess's law to determine the enthalpy of vaporization for water.
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