4.41 g of propane gas (c3h8) is injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below. the calorimeter (including the water) has a heat capacity of 97.1 kj/°c. c3h8(g) + 5 o2(g)  3 co2(g) + 4 h2o() (a) if the temperature rose from 25.000°c to 27.282°c, what is the heat of the reaction, qrxn?

Answer :  The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

where,

= heat released by the reaction = ?

= heat absorbed by the calorimeter

= specific heat of calorimeter =

= change in temperature =

Now put all the given values in the above formula, we get:

As,

So,

Thus, the heat of the reaction is -221.6 kJ

h let me think

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.