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ser-zykov [4K]
3 years ago
15

BRAINLIEST GIVE AWAY! Please give the correct answer I will reward you with Brainliest!

Chemistry
1 answer:
harkovskaia [24]3 years ago
7 0
Molar mass is the mass of each element multiplied by subscript, then added together.

So, we’ll do

C = 12g x 12 = 144 g
H = 1g x 22 = 22 g
O = 16g x 11 = 176 g

The total of all of these is the molar mass: 342 g in a mol

The amount of mols of each element in each compound is likely equal to its subscripts.

C = 12 mols
H = 22 mols
O = 11 mols

For the atoms, you’ll multiply each one of the above by Avogadro’s number

Avogadro’s Number: 6.02 x 10^23 atoms/mol


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Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is
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Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The equilibrium reaction follows:

2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = \frac{-9078.57}{2}=-4539.28kJ

For per gram of compound:

Molar mass of B_5H_9 = 63.12 g/mole

\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

7 0
3 years ago
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