The mass, in grams, of hydrogen that would be produced from the reaction will be 0.102 grams
Stoichiometric calculations
From the balanced equation of the reaction:
[tex]Mg + 2H_2O -- > Mg(OH)_2 + H_2[/tex]
The mole ration of magnesium to water is 1:2.
Mole of 4.73 grams Mg = 4.73/24.3 = 0.195 moles
Mole of 1.83 grams water = 1.83/18 = 0.102 moles
Hence, water is limiting.
Mole ratio of water to the hydrogen produced = 2:1
Equivalent mole of hydrogen gas produced = 0.102/2 = 0.05 moles
Mass of 0.05 moles hydrogen gas = 0.05 x 2 = 0.102 grams
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