Question

Given the unbalanced equation: ____ Al(s) + ____Cl2(g) → _____AlCl3(s) If 50.00 g of aluminum and 150.00 g of chlorine gas are reacted. What is the limiting reactant?a) Cl2 and AlCl3 b)Al and Cl2 c) AlCl3 d) Al e) Cl2

Answer 1

Molar mass of Al = 26.98 g/mol

mass(Al)= 50.0 g

use:

number of mol of Al,

n = mass of Al/molar mass of Al

=(50 g)/(26.98 g/mol)

= 1.853 mol

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 150.0 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(1.5*10^2 g)/(70.9 g/mol)

= 2.116 mol

Balanced chemical equation is:

2 Al + 3 Cl2 ---> 2 AlCl3 +

2 mol of Al reacts with 3 mol of Cl2

for 1.853 mol of Al, 2.78 mol of Cl2 is required

But we have 2.116 mol of Cl2

so, Cl2 is limiting reagent

Answer: e