Answer: The amount of heat released when 0.211 moles of
reacts is 554.8 kJ
Step-by-step explanation:
The chemical equation for the reaction of
with oxygen gas follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H_(rxn)=[(5* \Delta H_f_((B_2O_3(s))))+(9* \Delta H_f_((H_2O(l))))]-[(2* \Delta H_f_((B_5H_9(l))))+(12* \Delta H_f_((O_2(g))))]](https://img.qammunity.org/2020/formulas/chemistry/college/er0xvsfanmzxhm4hn4ap6xuavivf2emamb.png)
We are given:

Putting values in above equation, we get:
![\Delta H_(rxn)=[(2* (-1272))+(9* (-285.4))]-[(2* (73.2))+(12* (0))]\\\\\Delta H_(rxn)=-5259kJ](https://img.qammunity.org/2020/formulas/chemistry/college/aevq3fse7o7ljgrvz4op7ndf3oj5khl1sz.png)
To calculate the amount of heat released for the given amount of
, we use unitary method, we get:
When 2 moles of
reacts, the amount of heat released is 5259 kJ
So, when 0.211 moles of
will react, the amount of heat released will be =

Hence, the amount of heat released when 0.211 moles of
reacts is 554.8 kJ