Question

If 6.2 moles of Mg(s) are added to 9.3 moles of H2O(l) how many moles of H2(g) can be made? Mg(s) + 2 H2O(l) → Mg(OH)2 (aq) + H2(g)

Answer 1

To calculate how many moles of hydrogen gas is formed, first we have to find which reactant is limiting reagent.

Limiting reagent is the reactant that decides the amount of product formed.

Mg{s) + 2 H_{​​​​​​2}​​​​​O (l) ----------------> Mg(OH)₂ (aq) + H_{​​​​​​2} (g)

One mole of Mg requires 2 moles of water

6.2 moles of Mg requires how many moles of water.

Number of moles of water required

= (6.2 moles /1 mole )*2 moles

=12.4 moles of water required.

Since only 9.3 moles of water is taken, it is the limiting reagent.

2 moles of water on reaction with excess of Mg forms one mole of hydrogen.

9.3 moles of water on reaction with excess of Mg forms how many moles of hydrogen.

Number of moles of H_{​​​​​​2} is formed = 4.65 moles