Answer:
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
