A chemist reacted 0.05 moles of sodium with water to form 50cm3 of sodium hydroxide solution.

Na + H2O -- NaOH + H2 (unbalanced)
What mass of sodium was reacted?

2Na + 2H2O -- 2NaOH + H2

mols Na = grams Na/atomic mass Na.
You know atomic mass Na and mols Na, solve for grams Na. The 50 cc NaOH has nothing to do with the problem assuming that the water used was in excess.

4cc

To determine the mass of sodium that was reacted, we need to use the mole ratio between sodium and sodium hydroxide.

Given:
Moles of sodium hydroxide (NaOH) = 0.05 moles
Volume of sodium hydroxide solution = 50 cm^3

Let's first balance the chemical equation:

2Na + 2H2O -> 2NaOH + H2

From the balanced equation, the mole ratio between sodium (Na) and sodium hydroxide (NaOH) is 2:2 or 1:1.

Therefore, the moles of sodium (Na) reacted will be equal to the moles of sodium hydroxide (NaOH).

Moles of sodium (Na) = 0.05 moles

Now, let's calculate the molar mass of sodium (Na). The atomic mass of sodium is 22.99 g/mol.

Molar mass of sodium (Na) = 22.99 g/mol

Finally, we can calculate the mass of sodium reacted using the formula:

Mass = Moles x Molar Mass

Mass of sodium reacted = 0.05 moles x 22.99 g/mol

Mass of sodium reacted = 1.15 g

Therefore, the mass of sodium that was reacted is 1.15 grams.

To determine the mass of sodium that was reacted, you'll need to follow these steps:

Step 1: Find the molar mass of sodium (Na).
The molar mass of Na is approximately 22.99 g/mol.

Step 2: Calculate the number of moles of sodium used.
Given that 0.05 moles of sodium were reacted, no further calculation is needed for this step.

Step 3: Convert the number of moles of sodium to grams.
To convert moles to grams, use the formula:
mass (grams) = number of moles × molar mass
So, for the given reaction, the mass of sodium reacted would be:
mass (grams) = 0.05 moles × 22.99 g/mol

Calculating this:
mass (grams) = 1.1495 grams

Therefore, the mass of sodium that was reacted is approximately 1.1495 grams.