Answer:
4.44 g
Step-by-step explanation:
First, we need to determine the limiting reactant. The balanced chemical equation is:
2Na + 2H2O → 2NaOH + H2
From the equation, we see that 2 moles of sodium reacts with 2 moles of water to produce 1 mole of hydrogen.
Moles of Na in 120 g of Na:
Molar mass of Na = 22.99 g/mol
Number of moles of Na = mass/molar mass = 120 g / 22.99 g/mol = 5.22 mol
Moles of water in 80 g of H2O:
Molar mass of H2O = 18.02 g/mol
Number of moles of H2O = mass/molar mass = 80 g / 18.02 g/mol = 4.44 mol
Since 5.22 moles of Na requires 5.22 moles of H2O, water is the limiting reactant.
The moles of hydrogen produced is equal to the moles of the limiting reactant, which is 4.44 mol.
The mass of hydrogen produced is:
Mass = number of moles x molar mass
Mass = 4.44 mol x 1 g/mol = 4.44 g
So the grams of hydrogen produced by the 120 grams Na and 80 grams of water is 4.44 grams.