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Chapter 6: Problem 125

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 heat released when 54.0 g of diborane is combusted is -3973 kJ.

Step by step solution

01

Calculate the amount of moles of diborane

To calculate the amount of moles, we will use the formula: Moles = Mass (g) / Molar mass (g/mol) Diborane has a molar mass of (2 × 10.81) + (6 ×1.01) = 27.62 g/mol. So, Moles of diborane = 54.0 g / 27.62 g/mol
02

Find the moles of B₂O₃ and H₂O produced

From the balanced reaction: 1 mol B₂H₆ produces 1 mol B₂O₃ and 3 mol H₂O. But we calculated a different number of moles for B₂H₆ in Step 1, so using the stoichiometric ratios: Moles of B₂O₃ = moles of B₂H₆ produced Moles of H₂O = 3 × moles of B₂H₆ produced
03

Calculate the heat released

The enthalpy change provided is the heat released for the balanced equation. We can rewrite this as: Heat = ΔH × moles of B₂H₆ For this reaction, ΔH = -2035 kJ/mol, Plug in the values to calculate the heat released: Heat released = (-2035 kJ/mol) × moles of B₂H₆ from step 1
04

The final answer

The final result is the calculated heat released from the reaction, obtained by solving the equation in step 3: Heat released = (-2035 kJ/mol) × (54 g / 27.62 g/mol) Heat released = (-2035 kJ/mol) × (1.954 mol) Heat released = -3973 kJ Thus, the heat released when 54.0 g of diborane is combusted is -3973 kJ (negative sign indicating that the heat is released).

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Most popular questions from this chapter

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

For the following reactions at constant pressure, predict if $\Delta H>\Delta E, \Delta H<\Delta E,\( or \)\Delta H=\Delta E .$ a. \(2 \mathrm{HF}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) b. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$ c. $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$

Give the definition of the standard enthalpy of formation for a substance. Write separate reactions for the formation of \(\mathrm{NaCl}\) , \(\mathrm{H}_{2} \mathrm{O}, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) and \(\mathrm{PbSO}_{4}\) that have \(\Delta H^{\circ}\) values equal to $\Delta H_{\mathrm{f}}^{\circ}$ for each compound.

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 \(\mathrm{kJ} / \mathrm{mol.} )\)

Which has the greater kinetic energy, an object with a mass of 2.0 \(\mathrm{kg}\) and a velocity of 1.0 \(\mathrm{m} / \mathrm{s}\) or an object with a mass of 1.0 \(\mathrm{kg}\) and a velocity of 2.0 $\mathrm{m} / \mathrm{s}$ ?
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