Respuesta :
The reaction between metallic Magnesium and steam will be
[tex]Mg(s)+2H_{2}O(g)--->Mg(OH)_{2}+H_{2}(g)[/tex]
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
