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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

Sagot :

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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

Further explanation

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

  • Limiting reacttant

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

  • volume of H₂

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

  • mass of salt(AlCl₃)

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

  • mass of excess(Al) left

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

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