Question

Compounds of boron and hydrogen are remarkable for their unusual bonding and also for their reactivity. With the more reactive halogens, for example, diborane (B2H6) forms trihalides even at low temperatures: B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g) Δ Hrxn = −755.4 kJ How much heat is released when 4.465 kg of diborane reacts? (Give your answer in scientific notation.)

Answer 1

Answer:

Amount of energy released = $1.22\times 10^5\ kJ$

Explanation:

For $B_2H_6$  :-

Mass of $B_2H_6$  = 4.465 kg = 4465 g ( 1 kg = 1000 g)

Molar mass of $B_2H_6$  = 27.66 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{4465\ g}{27.66\ g/mol}$

$Moles\ of\ B_2H_6= 161.42\ mol$

From the reaction,

$B_2H_6_{(g)} + 6 Cl_2_{(g)}\rightarrow 2 BCl_3_{(g)}+ 6 HCl_{(g)} \Delta\ H_{rxn}=-755.4 kJ$

1 mole of $B_2H_6$ releases 755.4 kJ of energy

161.42 moles of $B_2H_6$ releases 755.4*161.42 kJ of energy

Amount of energy released = $1.22\times 10^5\ kJ$