Question

Ammonia reacts with oxygen according to the equation 4NH3(g)+5O2(g)→4NO(g)+6H2O(g),ΔHrxn=−906 kJ Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH3.

Answer 1

The heat associated with the complete reaction of 155 g of NH3 is -5133.1 kJ.

Step-by-step explanation:

The heat associated with the complete reaction of 155 g of NH3 can be calculated using the equation and given enthalpy value.

First, we need to convert the mass of NH3 to moles. The molar mass of NH3 is 17.03 g/mol, so:

155 g NH3 x (1 mol NH3/17.03 g NH3) = 9.09 mol NH3

Next, we can use the stoichiometry of the reaction to determine the moles of products:

9.09 mol NH3 x (4 mol NO/4 mol NH3) = 9.09 mol NO

9.09 mol NH3 x (6 mol H2O/4 mol NH3) = 13.64 mol H2O

Finally, we can calculate the heat using the enthalpy value:

9.09 mol NO x (-906 kJ/4 mol NO) = -2060.5 kJ

13.64 mol H2O x (-906 kJ/4 mol H2O) = -3072.6 kJ

Sum the heats of the products: -2060.5 kJ + -3072.6 kJ = -5133.1 kJ

Therefore, the heat associated with the complete reaction of 155 g of NH3 is -5133.1 kJ.

Answer 2

-956 kJ

Step-by-step explanation: