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spin [16.1K]
3 years ago
10

Calculate the number of grams of solid aluminium chloride that will form when a mixture containing 0.150 g of aluminum powder an

d 1.00 g of chlorine gas is allowed to react. 2al (s) + 3cl2 (g) ---> 2alcl3 (s)
Chemistry
1 answer:
IgorLugansk [536]3 years ago
7 0

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

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