CHEMISTRY | MOLE CONCEPT | Question 3
a) What mass of AL2O3 can be produced from the reaction of 10.00g of AL and 19.0g of O3?
b) How much of the excess reagent remains unreacted?
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DATA GIVEN:
Mass of Al = 10.00g
Mass of O3 = 19.0g
Asked:
Mass of Al2O3 produced.
Equation
2Al + O3 → Al2O3
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Mass of Al2O3 produced
Mass = mass /molar mass
= 10.00g /27g/mole
= 0.37 mole
Mole ratio = 0.37 mole /2 mole
= 0.185 mole
Mole of O3 = 19.0g /48g/mole
= 0.3958mole
Mole ratio for O3 = 0.3959 mole
Limiting is Al, excess is O3
Therefore, mole of Al2O3 produced
2mole of Al Ξ 1 mole 0f Al2O3
0.37 mole Ξ ??
= 0.37 mole × 1 mole /2 mole
= 0.1979 mole
Mass = 0.1979 mole × 102g mole
= 20.1858g
20.1858g of Al2O3 was produced -
Mass of O3 that remained.
Mole equivalent between Al & O3
2 mole Ξ 1 mole of O3
0.37 mole Ξ ?? of O3
= 0.37 mole × 1 mole /2 mole
= 0.185 mole of O3
Mass = 0.185 mole × 48g/mole
= 8.88g reacted for O3
Mass remain = 19.0g – 8.88g
= 10.12g
10,12g of excess reagent (03) was remained unreacted
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