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Option B. 4.5g

Explanation:

The balanced equation for the reaction is given below:

2Na + 2H2O —> 2NaOH + H2

Next, we shall determine the mass of Na and the mass of H2O that reacted and the mass of H2 produced from the balanced equation. This is illustrated below:

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar mass of H2O = (2.02x1) + 16 = 18.02g/mol

Mass of H2O from the balanced equation = 2 x 18.02 = 36.04g

Molar mass of H2 = 2.02 x 1 = 2.02g/mol.

Mass of H2 from the balanced equation = 1 x 2.02 = 2.02g

From the balanced equation above,

46g of Na reacted with 36.04g of H2O to produce 2.02g of H2.

Next, we shall determine the limiting reactant. This is illustrated below:

From the balanced equation above,

46g of Na reacted with 36.04g of H2O.

Therefore, 120g of Na will react with = (120 x 36.04)/46 = 94.02g of H2O.

From the calculations made above, we can see that it will take a high mass of H2O i.e 94.02g than what was given i.e 80g to react completely with 120g of Na. Therefore, H2O is the limiting reactant and Na is the excess reactant.

Finally, we shall determine the mass of H2 produced from the reaction as follow:

The limiting reactant will be used in this case as it will give the maximum yield of H2 since all of it is used up in the reaction.

From the balanced equation above, 36.04g of H2O reacted to produce 2.02g of H2.

Therefore, 80g of H2O will react to produce = (80 x 2.02)/36.04 = 4.5g of H2.

Therefore, 4.5g of H2 is produced from the reaction.