Calculate the number of grams of solid aluminium chloride that will form when a mixture containing 0.150 g of aluminum powder and 1.00 g of chlorine gas is allowed to react. 2al (s) + 3cl2 (g) > 2alcl3 (s)

0.630 g

Step-by-step explanation:  

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, lets assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    26.98   70.91      133.34  

           2Al   +   3Cl₂ ⟶ 2AlCl₃

m/g:  0.150       1.00

Step 1.  Calculate the moles of each reactant  

Moles of Al   = 0.150 g × 1 mol/26.98 g  = 0.005 560 mol

Moles of Cl₂ =  1.00  g  × 1 mol/70.91  g  = 0.014 10     mol

Step 2. Identify the limiting reactant

Calculate the moles of AlCl₃ we can obtain from each reactant.

From Al:  

The molar ratio of AlCl₃:Al is 2:2

Moles of AlCl₃ = 0.005 560 × 2/2  

Moles of AlCl₃ = 0.005 560 mol AlCl₃

From Cl₂:  

The molar ratio of AlCl₃: Cl₂ is 2:3.  

Moles of AlCl₃ = 0.014 10 × 2/3  

Moles of AlCl₃ = 0.009 401 mol AlCl₃  

Al is the limiting reactant because it gives the smaller amount of AlCl₃.  

Step 4. Calculate the theoretical yield.

Theor. yield = 0.005 560 mol AlCl₃ × 133.34 g AlCl₃/1 mol AlCl₃

Theor. yield = 0.630 g AlCl₃