Concept explainers
(a)
Interpretation:
The mass percentage of Mg in a magnesium-aluminium alloy that has a mass of 0.263 g is to be determined.
Concept introduction:
Mass percent is employed to determine the concentration of one compound in a mixture of the compound. The formula to calculate mass percent is as follows:
Mass percent of compound=(mass of the compoundmass of mixture)
(a)
Answer to Problem 4.137P
The mass percentage of Mg in a magnesium-aluminium alloy is 22.7 %.
Explanation of Solution
Consider the mass of Mg is x g and mass of Al is (0.263−x)g.
The formula to calculate the volume of the alloy is as follows:
Volume of alloy=(mass of alloydensity of alloy) (2)
Consider 0.263 g the mass of alloy and 2.40 g/cm3 for the density of alloy in the equation (2).
Volume of alloy=(0.263 g2.40 g/cm3)=0.10958 cm3
The formula to calculate the volume of Mg is as follows:
Volume of Mg=(mass of Mgdensity of Mg) (3)
Consider x g the mass of Mg and 1.74 g/cm3 for the density of Mg in the equation (3).
Volume of Mg=(x g1.74 g/cm3)
The formula to calculate the volume of Al is as follows:
Volume of Al=(mass of Aldensity of Al) (4)
Consider (0.263−x)g the mass of Al and 2.70 g/cm3 for the density of Al in the equation (4).
Volume of Al=((0.263−x)g2.70 g/cm3)
The formula to calculate x is as follows:
Volume of alloy=Volume of Mg+Volume of Al (5)
Substitute 0.10958 cm3 for the volume of alloy, (x g1.74 g/cm3) for the volume of Mg and ((0.263−x)g2.70 g/cm3) for the volume of Al in the equation (5).
0.10958 cm3=(x g1.74 g/cm3)+((0.263−x)g2.70 g/cm3)x=0.05957 g
The expression to calculate the mass percent of Mg is:
mass % of Mg=(mass of Mg(g)mass of alloy sample(g))(100) (6)
Substitute 0.05957 g for the mass of Mg and 0.263 g for the mass of alloy sample in the equation (6).
Mass % of Mg=(0.05957 g0.263 g)(100)=22.6502 %≈22.7 %
The mass percentage of Mg in a magnesium-aluminium alloy is 22.7 %.
(b)
Interpretation:
The mass percentage of Mg in a magnesium-aluminium alloy that reacts with excess aqueous HCl and forms 1.38×10−2 mol H2 is to be determined.
Concept introduction:
Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.
Consider the general reaction,
A+2B→3C
One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.
(b)
Answer to Problem 4.137P
The mass percentage of Mg in a magnesium-aluminium alloy is 21.6 %.
Explanation of Solution
The reaction of Mg and Al with HCl is as follows:
Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)
Consider the mass of Mg is x g and mass of Al is (0.263−x)g.
The formula to calculate moles of H2 from Mg is as follows:
moles of H2=(mass of Mgmolar mass of Mg)(1 mol H21 mol Mg) (7)
Substitute x g for the mass of Mg and 24.31 g/mol for molar mass of Mg in the equation (7).
moles of H2=(x g24.31 g)(1 mol H21 mol Mg)
The formula to calculate moles of H2 from Al is as follows:
moles of H2=(mass of Almolar mass of Al)(3 mol H22 mol Al) (8)
Substitute (0.263−x)g for the mass of Al and 26.98 g/mol for molar mass of Al in the equation (8)
moles of H2=((0.263−x)g26.98 g)(3 mol H22 mol Al)
The formula to calculate x is as follows:
moles of H2 produced=moles of H2 from Mg+moles of H2from Al (9)
Substitute (x g24.31 g/mol)(1 mol H21 mol Mg) for moles of H2 from Mg, 1.38×10−2 mol for moles of H2 produced and ((0.263−x)g26.98 g/mol)(3 mol H22 mol Al) for moles of H2 from Al in the equation (9).
1.38×10−2 mol=(x g24.31 g/mol)(1 mol H21 mol Mg)+((0.263−x)g26.98 g/mol)(3 mol H22 mol Al)8.22×10−4=0.014462xx=0.05684 g
Substitute 0.05684 g for the mass of Mg and 0.263 g for the mass of alloy sample in the equation (6).
Mass % of Mg=(0.05684 g0.263 g)(100)=21.6122 %≈21.6 %
The mass percentage of Mg in a magnesium-aluminium alloy is 21.6 %.
(c)
Interpretation:
The mass percentage of Mg in a magnesium-aluminium alloy that reacts with excess O2 and forms 0.483 g of oxide is to be determined.
Concept introduction:
Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.
Consider the general reaction,
A+2B→3C
One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.
(c)
Answer to Problem 4.137P
The mass percentage of Mg in a magnesium-aluminium alloy is 22.9 %.
Explanation of Solution
The reaction of Mg and Al with O2 is as follows:
Mg(s)+O2(aq)→2MgO(s)4Al(s)+3O2(aq)→2Al2O3(s)
Consider the mass of Mg is x g and mass of Al is (0.263−x)g.
The formula to calculate the mass of MgO from Mg is as follows:
mass of MgO=(mass of Mgmolar mass of Mg)(2 mol MgO2 mol Mg)(molar mass of MgO) (10)
Substitute x g for the mass of Mg, 24.31 g/mol for molar mass of Mg and 40.31 g/mol for the molar mass of MgO in the equation (10).
mass of MgO=(x g24.31 g/mol)(2 mol MgO2 mol Mg)(40.31 g/mol)
The formula to calculate the mass of Al2O3 from Al is as follows:
mass of Al2O3=(mass of Almolar mass of Al)(2 mol Al2O34 mol Al)(molar mass of Al2O3) (11)
Substitute (0.263−x)g for the mass of Al, 26.98 g/mol for molar mass of Al and 101.96 g/mol for the molar mass of Al2O3 in the equation (11).
mass of Al2O3=((0.263−x)g26.98 g/mol)(2 mol Al2O34 mol Al)(101.96 g/mol)
The formula to calculate x is as follows:
mass of oxide produced=mass of MgO from Mg+mass of Al2O3from Al (12)
Substitute (x g24.31 g/mol)(2 mol MgO2 mol Mg)(40.31 g/mol) for the mass of MgO from Mg, 0.483 g for moles of oxide produced and ((0.263−x)g26.98 g/mol)(2 mol Al2O34 mol Al)(101.96 g/mol) for moles of Al2O3 from Al in the equation (12).
0.483 g=[(x g24.31 g/mol)(2 mol MgO2 mol Mg)(40.31 g/mol)+((0.263−x)g26.98 g/mol)(2 mol Al2O34 mol Al)(101.96 g/mol)]0.01359=0.2315xx=0.060298 g
Substitute 0.060298 g for the mass of Mg and 0.263 g for the mass of alloy sample in the equation (6).
Mass % of Mg=(0.060298 g0.263 g)(100)=22.927 %≈22.9 %
The mass percentage of Mg in a magnesium-aluminium alloy is 22.9 %.
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