Magnesium 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?

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction is:

2 Mg(s) + I2(g) → 2 MgI(s)

Using the given masses of Mg and iodine, we can determine which one is limiting and calculate the theoretical yield of MgI2.

First, we need to convert the masses to moles, using their molar masses:

moles of Mg = 5.15 g / 24.31 g/mol = 0.212 mol
moles of I2 = 50.0 g / 253.81 g/mol = 0.197 mol

Since the stoichiometry of the equation is 2:1 (i.e. 2 moles of Mg react with 1 mole of I2), we can see that Mg is in excess, and iodine is limiting.

The theoretical yield of MgI2 can be calculated from the amount of limiting reagent (iodine) used:

moles of MgI2 = moles of I2 used (from the equation) = 0.197 mol
mass of MgI2 = moles of MgI2 x molar mass of MgI2
= 0.197 mol x (24.31 g/mol + 2x 126.9 g/mol) = 54.8 g

Therefore, the best choice is C, which gives a theoretical yield of 54.8 g MgI2.

Magnesium Reacts with Iodine Gas at High Temperatures to Form