Answer:
0.185moles of Al₂O₃
Explanation:
Mass of Al = 10g
Mass of O₂ = 19g
Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃
This is the balanced reaction equation.
Solution
From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.
Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.
Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:
Number of moles of Al = [tex]\frac{mass}{molar mass}[/tex]
= [tex]\frac{10}{27}[/tex]
= 0.37mol
From the equation:
4 moles of Al produced 2 moles of Al₂O₃
0.37 mole will yield: [tex]\frac{2 x 0.37}{4}[/tex] = 0.185moles of Al₂O₃
0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.
The equation of the reaction between aluminium and oxygen gas to produce aluminium oxide is given below as follows:
[tex]4Al + 3O₂ → 2Al₂O₃[/tex]
From the equation of reaction 4 moles of aluminium reacts with 3 moles of oxygen to produce 2 moles of aluminium oxide.
The moles of reactants is calculated using:
molar mass of aluminium = 27.0 g
molar mass of oxygen = 32.0 g
moles of aluminium = 10/27 = 0.37 moles
moles of oxygen = 19/32 = 0.59
Aluminium is the limiting reactant
Thus, moles of aluminium oxide produced = 0.37 × 2 = 0.74 moles
Therefore, 0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.
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