To conduct the synthesis of iodosalicylamide, Edward used 1.07 g of salicylamide (MW: 137.14 g/mol) and 1.68 g of sodium iodide (MW:149.89 g/mol). Assuming the reaction yield is 100%, how many grams of iodosalicylamide (MW:263.03 g/mol) would be formed

2.05 g

Explanation:

Number of moles of salicylamide = 1.07g/137.14 g/mol = 0.0078 moles

Number of moles of NaI = 1.68 g/149.89 g/mol = 0.011 moles

Since the reaction is in a ratio of 1:1, salicylamide is the limiting reactant and 0.0078 moles of iodosalicylamide is formed.

Hence, theoretical yield of iodosalicylamide = 0.0078 moles × 263.03 g/mol = 2.05 g

Since there is 100% yield, actual yield = theoretical yield = 2.05 g

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 :