A 4.98 g sample of aniline (C_6H_5NH_2) was combusted in a bomb calorimeter with

Question

A 4.98 g sample of aniline (C_6H_5NH_2) was combusted in a bomb calorimeter with a heat capacity of 4.25 kJ/degree C. If the temperature rose from 29.5 degree C to 69.8 degree C, determine the value of Delta H for the combustion aniline. [molar mass C_6H_5NH_2 = 93.13 g/mol] Calculate the energy (in kJ) transferred when 3.621 g of BaO is produced according to the following equation, [molar masses: Ba = 137.30 g/mole; O_2 = 32.00 g/mole; BaO = 153 30 o/mo| 2 Ba (s) + O_2(g) rightarrow 2 BaO (s) Delta H = -1107 kJ

Explanation / Answer

4) Q = delta H = C(T2-T1) C = heat capacity of calorimeter

Given that

mass of aniline = 4.98 g

Molar mass of aniline = 93.13 g/mol

Moles of the aniline = mass of compund/ Molar mass of compound

= 4.98 g/93.13 g/mol

= 0.053 mol

Hence,

delta H = nC(T2-T1) = 0.053 mol x 4.25 kJ/oC x (69.8 oC - 29.5 oC )

= 90.77 kJ

5) 2Ba + O2 ----------> 2BaO delta H = -1107 kJ

2 mol 1107 kJ heat is released

2 x 153.3 = 306.6 g   1107 kJ heat is released

3.621 g ?

? = (3.621 g/ 306.6 g ) x 1107 kJ heat is released

= 13.07 kJ heat will be released.

Therefore,

13.07 kJ energy will be transferred.

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