Question

Consider the following thermochemical reaction for kerosene:

Answer 1

We have to solve this question using the stoichiometry of the reaction:

The equation of the reaction is;

$2 C12H26(l) + 37 O2(g) -----> 24 CO2(g) + 26 H2O(l) + 15,026 kJ$

According to the question;

Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles

From the  stoichiometry of the reaction:

Since;

24 moles of CO2 released 15,026 KJ

0.48 moles of CO2 will release 0.48 * 15,026/24

= 301 KJ of heat.

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Answer 2

Answer:

2 C12H26(l) + 37 O2(g) ----- > 24 CO2(g) + 26 H2O(l) + 15,026 kJ2C12H26(l)+37O2(g)−−−−−>24CO2(g)+26H2O(l)+15,026kJ

According to the question;

Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles

Since;

24 moles of CO2 released 15,026 KJ

0.48 moles of CO2 will release 0.48 * 15,026/24

= 301 KJ of heat.