Question

12 gram mg react with dilute mineral acid to produce maximum hydrogen equal to

Answer 1

Mg + 2HCl = MgCl2 + H2

Mg + 2 H{+} → Mg{2+} + H2

Supposing the question to be "How much hydrogen is produced?":

(12 g Mg) / (24.30506 g Mg/mol) x (1 mol H2 / 1 mol Mg) = 0.49 mol H2

(0.49 mol H2) x (22.414 L/mol) = 11 L H2 at STP

(0.49 mol H2) x (2.015894 g H2/mol) = 0.99 g H2

aprox 1 gm of H2

OR

Mg + 2HCl = MgCl2 + H2

So One mole of Mg reacts with 2 moles of H2 and gives 1 mole of H2

Now, as atomic weight of Mg=24, no. of moles of Mg present = 0.5

as 1 mole of Mg reacts with 2 moles of HCl to give 1 mole H2

0.5 moles of Mg will react with 1 mole of HCl to give 0.5 moles of H2

0.5 moles of H2 will result into formation of 2x0.5= 1gm of hydrogen