Question

For the reaction shown, compute the theoretical yield of the product in grams for each of the following initial amounts of reactants. 2Al(s)+3Cl2(g)→2AlCl3(s)

A) 1.0 g Al; 1.0 g Cl₂

B) 5.5 g Al; 19.8 g Cl₂

C) 0.439 g Al; 2.29 g Cl₂

Answer 1

As the reaction is given by

2 Al(s) + 3 Cl2(g) --> 2 AlCl3 (s)
molar mass of Al = 27g/mol , molar mass of Cl = 35.5 g/mol , we get molar mass of Cl2 = 71 g/mol

Now we calculate the no of moles of the reactants

no of moles = given mass / molar mass

1) no of moles of Al = 1 g /27 g/mol = 0.037 moles

no of moles of Cl2 = 1 g /71/mol = 0.014 moles           

As , 3 moles of Cl2 react with 2 moles Al to produce 2 moles AlCl3

1 moles of CL2 reacts with 2/3 moles of Al                                           

0.014 moles of CL2 reacts with 0.014 * 0.0246 = 0.0093 moles of Al                

CL2 is limiting agent here and some of the AL remains unreacted .

hence yield of the product is

= no of moles * molar mass = 133.5 g /mol * 0.0093 mol = 1.241 g of AlCL3 is formed

2)  no of moles of Al =   5.5g / 27 g/mol = 0.203 moles

no of moles of Cl2 = 19.8g / 71g/mol = 0.278   moles           

As , 3 moles of Cl2 react with 2 moles Al to produce 2 moles AlCl3

1 moles of CL2 reacts with 2/3 moles of Al                                           

0.203 moles of CL2 reacts with 0.203 * 2/3 = 0.136 moles of Al                

CL2 is limiting agent here and some of the AL remains unreacted .

hence yield of the product is

= no of moles * molar mass = 133.5 g /mol * 0.136 mol = 18.156  g ALCL3 is formed

3)

no of moles of Al =   0.439 g / 27 g/mol = 0.0162  moles

no of moles of Cl2 = 2.29 g / 71g/mol = 0.0322   moles           

As , 2 moles of Al react with 3 moles CL2 to produce 2 moles AlCl3

1 moles of Al reacts with 2/3  moles of CL2                                           

0.0162   moles of AL reacts with     2/3 * 0.0162 moles of CL2 = 0.0108 moles of CL2

                

AL is limiting agent here and some of the CL2 remains unreacted .

hence yield of the product is

= no of moles * molar mass = 133.5 g /mol * 0.0162 mol = 2.162 g of ALCL3 is formed