Question

Metallic magnesium reacts with steam to produce magnesium hydroxide and hydrogen gas. If 19.4 of Mg are heated with 14g of H2O what is the limiting reaction?

How much is the excess reactant is left?
Answer in units of mol

How much Mg(OH)2 is
formed?
Answer in units of g

Answer 1

The reaction between metallic Magnesium and steam will be

`Mg(s)+2H_(2)O(g)--->Mg(OH)_(2)+H_(2)(g)`

As per balanced equation:

one mole of Mg will react with two moles of water to give one mole of magnesium hydroxide

Moles of Mg = Mass / atomic mass = 19.4/24 = 0.81

Moles of water = mass / molar mass = 14/18 =0.78

thus here for 0.81 moles of Mg we need twice the moles of water

However we have just 0.78 moles of water it means the limiting reagent is water

The moles of Mg reacted with water = 0.5 X 0.78 = 0.39 moles

moles of Mg left after reaction = 0.81-0.39= 0.42 moles

mass of Mg left (excess reagent) = moles X atomic mass = 0.42 X 24 = 10.08g

The moles of Mg(OH)2 formed = 0.39 moles

mass of Mg(OH)2 formed = moles X molar mass = 0.39 X 58 = 22.62g