Magnesium reacts with iron(III) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants) and iron.
3Mg(s) + 2FeCl₃(s) $\rightarrow$ 3MgCl₂(s) + 2Fe(s)
A mixture of 41.0 g of magnesium (? = 24.31 g/mol) and 175 g of iron(III) chloride (? = 162.2 g/mol) is allowed to react. What mass of magnesium chloride = 95.21 g/mol) is formed?

Question

Magnesium reacts with iron(III) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants) and iron.
3Mg(s) + 2FeCl₃(s)

\rightarrow

3MgCl₂(s) + 2Fe(s)
A mixture of 41.0 g of magnesium (? = 24.31 g/mol) and 175 g of iron(III) chloride (? = 162.2 g/mol) is allowed to react. What mass of magnesium chloride = 95.21 g/mol) is formed?

Quizplus · Accepted Answer

The limiting reactant is iron(III) chloride. Calculate moles of each reactant: Mg = 41.0 g / 24.31 g/mol = 1.687 mol, FeCl3 = 175 g / 162.2 g/mol = 1.079 mol. The reaction requires 3 moles of Mg for every 2 moles of FeCl3, so FeCl3 is the limiting reactant. From the stoichiometry, 2 moles of FeCl3 produce 3 moles of MgCl2. Therefore, 1.079 moles of FeCl3 produce (3/2) * 1.079 = 1.619 moles of MgCl2. The mass of MgCl2 formed is 1.619 moles * 95.21 g/mol = 154 g.

Magnesium Reacts with Iron(III) Chloride to Form Magnesium Chloride (Which →\rightarrow→

Magnesium Reacts with Iron(III) Chloride to Form Magnesium Chloride (Which $\rightarrow$