Answer:
0.630 g
Step-by-step explanation:
We are given the mass of two reactants, so this is a <em>limiting reactant problem</em>.
We know that we will need mases, moles, and molar masses, so, lets <em>assemble all the data in one place</em>, 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 <em>moles of each reactant </em>
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 l<em>imiting reactant </em>
Calculate the moles of AlCl₃ we can obtain from each reactant.
<em>From Al</em>:
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
<em>From Cl₂:</em>
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 <em>limiting reactant </em>because it gives the smaller amount of AlCl₃.
<em>Step 4</em>. Calculate the <em>theoretical yield.</em>
Theor. yield = 0.005 560 mol AlCl₃ × 133.34 g AlCl₃/1 mol AlCl₃
Theor. yield = 0.630 g AlCl₃
