Consider the following balanced equation:

2 Na + 2 H2O ? 2 NaOH + H2
If 92.0 g of sodium reacts with 76.0 g of water until completion, which reactant will remain, and in what quantity?

To solve this problem, we need to determine which reactant is in excess and calculate the quantity of the remaining reactant.

First, we need to calculate the number of moles for each reactant using their respective molar masses.

The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of water (H2O) is 18.02 g/mol.

To calculate the number of moles:

Moles of sodium (Na) = mass of sodium (g) / molar mass of sodium (g/mol)
Moles of sodium (Na) = 92.0 g / 22.99 g/mol = 4.00 mol

Moles of water (H2O) = mass of water (g) / molar mass of water (g/mol)
Moles of water (H2O) = 76.0 g / 18.02 g/mol = 4.22 mol

Now, using the balanced equation, we can determine the stoichiometric ratio between sodium and water. From the equation, we can see that 2 moles of sodium react with 2 moles of water.

Comparing the moles calculated above, we see that there are an equal number of moles of sodium and water. This means that neither reactant is in excess, and both will be completely consumed during the reaction.

Therefore, there won't be any remaining reactant after the reaction is completed. All 92.0 g of sodium and 76.0 g of water will be converted into 4.00 moles of sodium hydroxide (NaOH) and 2.00 moles of hydrogen gas (H2) according to the balanced equation.