Question

Magnesium metal reacts with iodine gas at high temperatures to form magnesium iodide. what mass of mgi2 can be produced from the reaction of 5.15 g mg and 50.0 g i2

Answer 1

54.8 g of MgI2 can be produced. To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium Atomic weight of Iodine = 126.90447 Atomic weight of Magnesium = 24.305 Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394 Now determine how many moles of Iodine and Magnesium you have moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent. So figure out how many moles of magnesium will be consumed by the iodine 0.393997154 mole / 2 = 0.196998577 mole. This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2 0.196998577 mole * 278.11394 g/mole = 54.78805 g Round the result to the correct number of significant figures. 54.78805 g = 54.8 g