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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.