A student prepares iodoform from 2.90 g of propanone (58 g/mol) using iodine, 38.1 g (254 g/mol)…

A student prepares iodoform from 2.90 g of propanone ($58\ \text{g/mol}$) using iodine, 38.1 g ($254\ \text{g/mol}$), dissolved in potassium iodide solution, and sodium hydroxide solution added until the iodine colour fades. Pale yellow crystals form on gentle warming. After washing and drying, the mass is 14.2 g. Use H = 1, O = 16 and Na = 23 in g/mol, so that sodium hydroxide is $40\ \text{g/mol}$, and take iodoform as $394\ \text{g/mol}$. The reagents are exactly in the ratio of the equation. Find the amount of iodine needed, the mass of sodium hydroxide needed, the theoretical mass of iodoform, and the percentage yield.