5.4 g of Aluminium reacts with 300 mL of 0.2 mol/L hydrochloric acid solution. a. Write equation for the reaction taking place. b. Specify which reactant is limiting and which reactant is excess. c. Find volume of the gas collected at S. T.P d. How many grams of salt are produced at the end of the reaction? e. How many grams of the excess reactant are left ate the end of the reaction? Given: Al=27 , H=1 , Cl=35.5

HCl as a limiting reactant

volume of the gas(H₂)= 0.672 L

mass of salt : =2.67 g

mass of excess left :  4.86 g

Given

5.4 g Al

300 mL of 0.2 mol/L HCl

Required

limitng reactants

volume of the gas(H₂)

mass of salt

mass of excess left

Solution

Reaction

2Al + 6HCl → 2AlCl₃ + 3H₂

mol Al = 5.4 g : 27 g/mol = 0.2

mol HCl = 0.3 L x 0.2 mol/L = 0.06

mol : coefficient of reactants :

Al = 0.2 : 2 = 0.1

HCl = 0.06 : 6 = 0.01

HCl as a limiting reactant(smaller ratio)

Al as an excess reactant

mol H₂ = 3/6 x mol HCl = 3/6 x 0.06 = 0.03

volume (STP : 1 mol=22.4 L) :

= 0.03 x 22.4 L

= 0.672 L

mol AlCl₃ = 2/6 x mol HCl = 2/6 x 0.06 = 0.02

mass AlCl₃ :

= mol x MW

= 0.02 x 133.5

=2.67 g

mol (reacted) : 2/6 x mol HCl = 2/6 x 0.06 = 0.02

mol (unreacted) : 0.2 - 0.02 = 0.18

mass = 0.18 x 27 g/mol = 4.86 g

explanation: