Answer:
1. 2.1 moles of Mg
2. 0.72 mole of Mg(OH)2
Explanation:
1. We'll begin by writing the balanced equation for the reaction. This is given below:
3Mg + 2AlBr3 —> 3MgBr2 + 2Al
From the balanced equation above, 3 moles of Mg reacted to produce 2 moles of Al.
Therefore, Xmol of Mg will react to produce 1.4 moles of Al i.e
Xmol of Mg = (3 x 1.4)/2
Xmol of Mg = 2.1 moles.
Therefore, 2.1 moles of Mg is required to 1.4 moles of Al.
2. We'll begin by calculating the number of mole in 26g of water, H2O.
This is illustrated below:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O = 26g
Number of mole of H2O =?
Mole = Mass /Molar Mass
Number of mole of H2O = 26/18
Number of mole of H2O = 1.44 moles
Next, we shall write the balanced equation for the reaction. This is given below:
2HNO3 + Mg(OH)2 —> Mg(NO3)2 + 2H2O
Finally, we can obtain the number of mole of Mg(OH)2 used in the reaction as follow:
From the balanced equation above,
1 mole of Mg(OH)2 reacted to produce 2 mole of H2O.
Therefore, Xmol of Mg(OH)2 will react to produce 1.44 moles of H2O i.e
Xmol of Mg(OH)2 = (1 x 1.44)/2
Xmol of Mg(OH)2 = 0.72 mole.
Therefore, 0.72 mole of Mg(OH)2 was used in the reaction.
