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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

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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

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