2. Sodium metal reacts with water to form sodium hydroxide and hydrogen gas
a. Balanced Chemical Reaction:
b. If you have 90.0 g of sodium and 80.0 g of water which is the limiting reactant and which is the excess reactant?
c. How many grams of the excess reactant are left over when the reaction is completed?
d. How many grams of hydrogen gas are produced?
1) 2Na + 2H2O = 2NaOH + H2
2) n = m / M
M (Na) = 23 g/mol
M (H2O) = 18 g/mol
M (H2) = 2 g/mol
n (Na) = 90 / 23 = 3.9 mol
n (H2O) = 80 / 18 = 4.4 mol
According to the equation n (Na) = n (H2O).
With the given amont of reactants Na is the limiting reactant.
3) The excess amount of H2O is: 4.4 - 3.9 = 0.5 mol.
m (H2O) = n x M = 0.5 x 18 = 9 g
4) According to the eqation, n (H2) = 1/2 x n (Na) = 3.9 / 2 = 1.95 mol
m (H2) = n x M = 1.95 x 2 = 3.9 g
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