Answer:
Six electrons are transferred in the formation of Al₂O₃.
Explanation:
Aluminium metal and Oxygen react to form Al₂O₃ as,
2 Al + 3/2 O₂ → Al₂O₃
Oxidation number of Al on left hand side is zero, while than on right hand side in Al₂O₃ is +3. Means it has lost 3 electrons per one atom and six electrons per two atoms. Also, the oxidation number of O at left hand side in O₂ is zero, while that in Al₂O₃ it is -2 per atom and -6 per 3 atoms.
So, two Al atoms have lost 6 electrons and 3 O atoms have gained six electrons.
A chemical reagent that is used in this experiment is silver nitrate (AgNO3). It is used to distinguish calcium chloride and calcium carbonate. when this reagent is used, silver from silver nitrate reacts with Chloride to calcium chloride and forms silver chloride, making a precipitates of white color.
1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.
Let's consider the reaction for the combustion of Mg.
Mg + 1/2 O₂ ⇒ MgO
1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

We can calculate the mass percent of O in MgO using the following expression.

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