Stoichiometry

Q1: If 1.5 moles of oxygen combines with aluminium to form Al2O3, then the weight of aluminium (atomic weight = 27 ) used in this reaction is __________ gm.

A 27

B 54

C 5.4

D 2.7

ANS:B - 54

To determine the weight of aluminum used in the reaction, we first need to find the molar ratio between aluminum and oxygen in the reaction. The balanced chemical equation for the formation of aluminum oxide (Al₂O₃) from aluminum (Al) and oxygen (O₂) is: 4𝐴𝑙+3𝑂2→2𝐴𝑙2𝑂34Al+3O2​→2Al2​O3​ From the balanced equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Given that 1.5 moles of oxygen are used in the reaction, we can calculate how many moles of aluminum are required by using the mole ratio: Moles of aluminum=1.5 moles of oxygen×4 moles of aluminum3 moles of oxygenMoles of aluminum=3 moles of oxygen1.5 moles of oxygen×4 moles of aluminum​ Moles of aluminum=2 molesMoles of aluminum=2 moles Now, to find the weight of aluminum, we use the formula: Weight=Number of moles×Molar massWeight=Number of moles×Molar mass Weight of aluminum=2 moles×27 g/molWeight of aluminum=2 moles×27 g/mol Weight of aluminum=54 gramsWeight of aluminum=54 grams So, the weight of aluminum used in this reaction is 54 g54g. Therefore, the correct answer is 54 g54g.



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