Question

A mixture of 8-hydroxyquinoline complexes of aluminum and magnesium weigh 1.0843 g. When ignited in a furnace open to the air, the mixutre decomposed, leaving a residue of Al2O3 and Mg weighing 0.1344. What is the weight percent of Al(C9H6NO3) in the original mixture? AlQ3 + MgQ2 → Al2O3 + MgO

Where: Q represents 8-hydroxyquinoline

AlQ3 F.W. = 459.441 g

MgQ2 F.W. = 312.611 g

Al2O3 F.W. = 101.961 g

MgO F.W. = 40.304 g

Answer #1

**consider the given equation**

**AlQ3 + MgQ2 = Al203 + MgO**

**let**

**weight of AlQ3 be z**

**now**

**moles = mass / molar mass**

**so**

**moles of AlQ3 = z / 459.441**

**now**

**weight of MgQ2 = 1.0843 - z**

**so**

**moles of MgQ2 = (1.0843 - z ) / 312.611**

**now**

**AlQ3 + MgQ2 = Al203 + MgO**

**we can see that**

**moles of Al203 = 0.5 x moles of AlQ3**

**so**

**moles of All203 = 0.5 x z / 459.441**

**so**

**mass of Al203 = 0.5 x z x 101.961 / 459.441 =
0.110962z**

**now**

**moles of MgO = moles of MgQ2 = (1.0843 - z ) /
312.611**

**so**

**mass of MgO = 40.304 x (1.0843 - z ) / 312.611 =
0.1397955 - 0.128927z**

**now**

**mass of Al203 + mass of MgO = 0.1344**

**so**

**0.110962z + 0.1397955 - 0.128927z = 0.1344**

**z = 0.3**

**now**

**we know that**

**% weight = mass of AlQ3 x 100 / total mass**

**so**

**% weight = 0.3 x 100 / 1.0843**

**% weight = 27.7**

**so**

**the weight percent of AlQ3 is 27.7 %**

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