Question

Consider the reaction of 1.35g aluminum and 100. ml of 1.20 M HBr

2Al(s) + 6HBr(aq)---->2AlBr3(aq) + 3 H2(g)

Employ an Ice Table in calculating these

find what amount of excess reagent will remain unreacted

Find the mass of AlBr3 should be collected if the reactions run at a 75% yield

Answer 1

The balanced reaction with ICE TABLE

Initial ICE TABLE

2Al(s) + 6HBr(aq)---->2AlBr3(aq) + 3 H2(g)

I 0.05 0.120

C - 2x - 6x +2x +3x

E (0.05-2x) (0.120-6x) 2x 3x

Moles of Al = mass/molecular weight

= (1.35 g) / (26.982 g/mol)

= 0.0500 moles

Moles of HBr = molarity x volume

= 1.20 mol/L x 100 mL x 1L/1000 mL

= 0.120 mol

From the stoichiometry of the reaction

2 mol Al required = 6 mol HBr

0.0500 mol Al required = 6*0.050/2 = 0.150 mol HBr

But we have less moles (0.120 moles) of HBr than required (0.150 mol)

HBr is limiting reactant

Al is excess reactant

Moles of Al unreached = Initial moles - reacted moles

= 0.0500 - 0.120*2/6

= 0.0500 - 0.0400

= 0.0100 mol

Mass if Al unreacted = moles x molecular weight

= 0.0100 mol x 26.982 g/mol

= 0.269 g

Part b

Moles of AlBr3 can be produced

= 2 mol AlBr3 x 0.120 mol HBr / 6 mol HBr

= 0.0400 mol

Theoretical yield of AlBr3 = moles x molecular weight

= 0.0400 mol x 266.69 g/mol

= 10.67 g

% yield = actual yield * 100 / theoretical yield

0.75 * 10.67 = actual yield

actual yield of AlBr3 = 8.00 g