2Mg + I2 --> 2MgI
0.555 g Mg (2 mol MgI/2 mol Mg)(151.21 g MgI/1 mol MgI)
= 83.9 grams magnesium iodide
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What mass of magnesium iodide is produced from the reaction of 0.555 g Mg with excess iodine?
2Mg + I2 --> 2MgI
0.555 g Mg (2 mol MgI/2 mol Mg)(151.21 g MgI/1 mol MgI)
= 83.9 grams magnesium iodide
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