Short Answers : Show all of your work(calcuations): 1. 60 grams of diatomic nitr

Question

Short Answers : Show all of your work(calcuations):

1. 60 grams of diatomic nitrogen are burned in the presence of oxygen to form nitrogen dioxide. How much (mass) oxygen was used? How much (mass) nitrogen dioxide was profuced?

2. 80 grams of propance (C3 H8) was burnt in a colorimeter. The calorimeter contained 200 grams of water at a temperature of 25 celsius before the reaction. After the reaction the temperature of the water was 30 celsius. What is the molar heat of combustion? The specific heat capacity of water is 4.184 Joules per grams per degree celsius.

3. Which of the following fuels woud produce the largest amount of heat energy: 20 grams of ethane (C2 H6) or 20 grams of ethanol (C2 H5 OH)? The heat of combustion for ethane os -1,560 KJ/mol and the heat of combustion for ethanol is -1,368 KJ/mol.

Explanation / Answer

Ans. #1. Balanced reaction:            N2(g) + 2O2(g) ----------> 2NO2(g)

Stoichiometry: 1 mol N2 reacts with 2 mol O2 to form 2 mol 2NO2.

Moles of N2 consumed = Mass/ molar mass = 60.0 g / (28.01348 g/mol) = 2.142 mol

Moles of O2 consumed = 2 x Moles of N2 consumed = 2 x 2.142 mol = 4.284 mol

Mass of O2 consumed = Moles x Molar mass = 4.284 mol x (31.9988 g/mol) = 137.08g

Moles of NO2 formed = 2 x Moles of N2 consumed = 2 x 2.142 mol = 4.284 mol

Mass of NO2 formed = Moles x Molar mass = 4.284 mol x (46.00554 g/mol) = 197.09g

#2. Total heat gained by water, q = m x s x dT             - equation 1   

Where,

m = mass in gram,

c = specific heat of water = 4.18 J g-10C-1

dT = change in temperature = (final – initial) temperature

            or, q = 200 g x 4.184 J g-10C-1 x 50C

                        = 4184 J

                        = 4.184 kJ

Assuming all heat released during propane combustion is absorbed by water, the total heat absorbed by water Is equal to total heat released during combustion.

So, 2.80 g propane produces 4.184 kJ energy.

Moles of propane combusted = Mass/ Molar mass

= 2.80 g/ (44.09652 g/mol) = 0.0635 mol

Molar heat of combustion = Amount of heat released / Moles of propane combusted

                                                = 4.184 kJ/ 0.0635 mol

                                                = 65.89 kJ/ mol

#3. A. Moles of ethane in 20 g sample = 20 g / (30 g/mol) = 0.6667 mol

Amount of heat released = Moles of ethane x molar heat of combustion

                                                = 0.6667 mol x (- 1560 kJ/mol)

                                                = - 1040.052 kJ

B. Moles of ethane in 20 g sample = 20 g / (46.0 g/mol) = 0.4348 mol

Amount of heat released = Moles of ethane x molar heat of combustion

                                                = 0.4348 mol x (- 1368 kJ/mol)

                                                = - 594.81 kJ

Therefore, ethane produces largest amount of heat.

Note that the -ve sign simply indicates that heat being released during combustion.

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