Answer:
(2) Cl₂
Explanation:
To find out which reactant is limiting, we need to compare the moles of each reactant to the stoichiometry of the balanced equation.
\(\sf \: Mg + Cl_2 \longrightarrow MgCl_2\)
The given reaction shows that 1 mole of Mg reacts with 1 mole of Cl₂
Molar Mass of Mg = 24.3 g/mol
Molar mass of Cl₂ = 70.9 g/mol
Let's calculate the Number of moles of each reactant.
\({\sf {Moles \ of \: Mg}} = \dfrac{20.0 \ {\sf{g}}}{24.3 \ { \sf{ g/mol}}} \\ \\ \ \ \ \ \ \ \ \ = 0.823 \: \sf{ moles}\)
\({\sf {Moles \ of \: Cl_2}} = \dfrac{50.0 \ {\sf{g}}}{70.9 \ { \sf{ g/mol}}} \\ \\ \ \ \ \ \ \ \ \ = 0.706 \: \sf{ moles}\)
Now we need to compare the mole ratio of mg to Cl₂ in the balanced equation, on comparing we get:
0.706 moles of Cl₂ is less than that of 0.823 moles of Mg. Hence, Cl₂ is the limiting reactant in the given reaction.
