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sukhopar [10]
3 years ago
14

Compounds of boron and hydrogen are remarkable for their unusual bonding and also for their reactivity. With the more reactive h

alogens, 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.)
Chemistry
1 answer:
sukhopar [10]3 years ago
7 0

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

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