Question

1- Using the balanced equation given below, calculate the number of grams of Br 2 needed to react with exactly 53.8 g of Al.

2Al (s) + 3Br2 = 2AlBr3(s)

2- Calculate the number of moles of CO2 generated by the reaction given below when 6.78 g of CaCO3 is heated.

CaCO3 = CaO + CO2(g)

3- Calculate the percent yield of the product if the actual yield is 22.45 g of AlBr3

Answer 1

Q1)

2Al + 3 Br2 ------------> 2AlBr3

53.8/27 = moles of al

thus 2 moles of Al requires 3 moles of Br2 to react completely

53.8/27 mol of Al requires = (53.8/27) x 3/2

= 2.99 mol of Br2

Thus mass of Br2 required = mol x molar mass

= 2.99x 159.8 g/mol

= 477.62 g of Br2

Q2) The reaction is

CaCO3 ----------------> CaO + CO2

1 mol of CaCO3 gives 1 mol of CO2

6.78/100 mol of CaCO3 gives = (6.78/100) mol x 1mol/1mol

= 6.78x10^-2 mol

3)

2Al + 3 Br2 ------------> 2AlBr3

53.8/27 = moles of Al are used

Thus 2 mol of AL gives 2 mol of AlBr3

Hence 53.8/27 mol of Al gives 53.8/27 mol of ALBr3

MAss of ALBr3 that can be formed = mol x molar mass

= (53.8/27) x 266.7 g/mol

= 531.42 g

Theoretical yield = 531.42 g of AlBr3

% yield = experimental yield x100/ theoretical yield

= 22.45 x 100/531.42

= 4.22%