Question

Magnesium reacts with iodine gas at high temperatures to form magnesium iodide. What mass of MgI₂ can be produced from the reaction of unlimited Mg and 13.4 g I₂?

Mg (s) + I₂ (l) $LaTeX: \rightarrow$ →MgI₂ (s)

Answer 1

Mg (s) + I₂ (l) ----> MgI₂ (s)

moles of I₂ = maas/molar mass

molar mass of I₂ = 253.8 g/mol

moles of I₂ = 13.4/253.8

                     = 0.0528 moles

from balanced equation:

1 mole of I₂ gives ----> 1 mole of MgI₂

So, moles of MgI₂ formed = moles of I₂ in the reaction.

moles of MgI₂ formed = 0.0528 moles

mass of MgI₂ = moles*molarmass             ( molarmass of MgI₂ = 278.1 g/mol)

                        = 0.0528*278.1

                        = 14.68 g