Question

How many grams of magnesium nitrate should be heated to produce oxygen which is just sufficient to form 15.3g of alumina?

Answer 1

1. Mole is defined as the amount of substance that contains the same number of units (atoms, molecules, or ions) as there are atoms in exactly 12g of a C-12 isotope
2. A number of moles can be calculated by dividing a given mass by molecular mass/ Atomic mass.
3. The equation for the decomposition of magnesium nitrate (Mg(NO₃)₂) is $2Mg{\left(N{O}_3\right)}_2\left(s\right)\stackrel{}{\to 2MgO\left(s\right)+4N{O}_2\left(g\right)+O_2}\left(g\right)$
4. 2 moles of magnesium nitrate Mg(NO₃)₂ produce 1 mole of oxygen.
5. Alumina (AI₂O₃) is formed by the reaction $4Al\left(s\right)+3O_2\left(g\right)\to 2A{I}_2{O}_3\left(s\right)$ 2 moles of Alumina is produced from 3 moles of Oxygen.
6. Number of moles of alumina in 15.3g of AI₂O3=$\frac{Givenmass}{Molecularmass}=\frac{15.3}{102}=0.15mole.$
7. No. of moles of oxygen to produce 0.15mole of alumina =$\frac{3}{2}x0.15=0.225mole$
8. No. of moles of magnesium nitrate needed to produce 0.225 mole of oxygen = 2x0.225 = 0.45 mole
9. Mass of magnesium nitrate in 0.45 moles

Mass of 1 mole of Mg(NO₃)₂ =148 gm

Mass of 0.45 moles of Mg(NO₃)₂=148x0.45 = 66.6g

Therefore, the mass of magnesium nitrate heated to produce oxygen which is sufficient for 15.3 g of alumina is 66.6g.