Part A

If the theoretical yield of the reaction below corresponds to 54.0 g and the actual yield was 52.0 g, calculate the percent yield.

Given: Li, 0 + 1,0 2LiOH

96.3%

Explanation:

Data obtained from the question include:

Theoretical yield = 54g

Actual yield = 52g

Percentage yield =..?

The percentage yield can be obtain as follow:

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 52/54 x 100

Percentage yield = 96.3%

Therefore, the percentage yield of the reaction is 96.3%

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :