Question

Q8:
What mass in grams of hydrogen is produced by the reaction of 4.73 g of magnesium with 1.83 g of water? (Mg =24, H = 1, O = 16)

Mg (s) + 2H2O (l) → Mg(OH)2 (s) + H2 (g)

a) 0.204
b) 0.051
c) 0.0162
d) 0.219

Answer 1

Mg + 2 H2O --> Mg(OH)2 + H2

MW of Mg = 24.3 g/mol

n of Mg = mass/MW = 4.73g / 24.3 g/mol = 0.195 mol

MW water H2O = 18 g/mol

n water = mass/MW = 1.83g / 18 g/mol = 1.017 mol

The amount of Magnesium is smaller than water, therefore allmagnesium will react with water.

but some water will remain in the reaction because there is no moremagnesium left to react with.

Therefore, the number of moles of magnesium hydroxide Mg(OH)2 andnumber of moles of hydrogen H2 are equal to number of moles ofmagnesium.

n H2 = n Mg = 0.195 mol

mass H2 = n*MW = 0.195*2 = 0.39 gram

Answer 2

Answer:

a) 0.204

Explanation:

First, you make a ratio between the reactants (H2O & Mg) to determine the limiting reactant so 2 mol H2O/ 1 mol Mg = 2

_calculate moles of Mg (mass/molar mass) = (4.73/24)= 0.197 moles

_calculate moles of H2O (mass/molar mass) = (1.83/18)= 0.1017 moles

make the ratio again between H2O & Mg= (0.1017/0.197)= 0.5

So, H2O is the limiting reactant and as you need the mass of H2 by comparing its moles with moles of  H2O, you will get 0.1017 moles

Finally, convert moles to grams by multiply it by the molar mass (0.1017 * 2)= 0.204 grams