A mixture containing 10.8\text{ g} of aluminium powder is reacted with 16.0\text{ g} of ox… — JAMB UTME

A mixture containing

10.8\text{ g}

of aluminium powder is reacted with

16.0\text{ g}

of oxygen gas according to the balanced chemical equation:

4\text{Al}_{(s)} + 3\text{O}_{2(g)} \rightarrow 2\text{Al}_2\text{O}_{3(s)}

What is the mass in grams of the excess reactant remaining unreacted at the end of the reaction?

[\text{Al} = 27, \text{O} = 16]

Cevap: 6.4 g

Cevap

The mass of the excess reactant (oxygen gas) remaining unreacted is 6.4 g.

To determine the unreacted mass of the excess reactant, first convert given masses to moles: 10.8 g of Al corresponds to 0.40 mol and 16.0 g of O₂ corresponds to 0.50 mol. Using the mole ratio from the balanced equation (4 moles Al : 3 moles O₂), 0.40 mol of Al reacts completely with 0.30 mol of O₂. Thus, Al is the limiting reactant and O₂ is in excess. The unreacted amount of O₂ is 0.50 mol - 0.30 mol = 0.20 mol. Converting 0.20 mol of O₂ back to mass using its molar mass of 32 g/mol yields 6.4 g.

Adım Adım Çözüm

Calculate the mole amounts of reactants provided

n(Al) = 0.40 mol, n(O₂) = 0.50 mol

Converting masses to moles using molar masses (Al = 27 g/mol, O₂ = 32 g/mol) is necessary for stoichiometric comparison.

Determine the theoretical moles of oxygen needed to react with all aluminium

n(O₂) required = 0.30 mol

From the balanced equation, 4 moles of Al require 3 moles of O₂, so 0.40 mol Al requires 0.40 × (3/4) = 0.30 mol O₂.

Identify the excess reactant and compute remaining moles

O₂ is in excess by 0.20 mol

Available O₂ (0.50 mol) exceeds required O₂ (0.30 mol), leaving 0.50 - 0.30 = 0.20 mol of O₂ unreacted.

Convert remaining moles of excess reactant back to mass

Mass of excess O₂ = 6.4 g

Multiplying 0.20 mol by the molar mass of O₂ (32 g/mol) gives the unreacted mass of oxygen.

Anahtar Kavram

Limiting and excess reactant calculations based on stoichiometric coefficients and mole conversions

Tahmini Süre:2m 0s

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