What mass in grams of sodium hydroxide is produced if 20.0 g of sodium meal reactts with excess water according to the chemical equation 2 Na(s)+2 H2O(l)=2 NaOH(aq)+H2(g)?

Set up an "ICE" table below the chemical equation with I being initial moles, C being change in moles, and E being end moles. Starting with I, 20.0 grams of sodium metal (20.0g x 1mol/22.990g of Na) is approximately .870 moles of sodium. We know from the equation that for every 2 moles of sodium metal reactant we get 2 moles of Sodium Hydroxide product out. So, .870 mols of Na x (2mols NaOH/2mols Na) is still .870 moles of sodium hydroxide. Now we convert back from moles to grams. .870 moles of Sodium hydroxide multiplied by it's molar mass which is 35.997grams per mole (22.990grams/mol Sodium + 15.999grams/mole Oxygen + 1.008grams/mole Hydrogen), is equal to 34.8grams of Sodium Hydroxide produced.

I think you're supposed to use stoichiometry for this problem. This is how I would do it: 20.0g Na x (1 mol Na / molar weight Na) x (2 mols NaOH / 2 mols Na) x (molar weight NaOH / 1 mol NaOH) = ____g NaOH.

To determine the mass of sodium hydroxide produced, we need to use the molar mass of sodium hydroxide (NaOH) and stoichiometry.

Step 1: Find the molar mass of NaOH.
The molar mass of an element or compound is obtained by summing up the atomic masses of all the atoms in it. The atomic masses are found on the periodic table.

Na: Atomic mass = 22.99 g/mol
O: Atomic mass = 16.00 g/mol
H: Atomic mass = 1.01 g/mol

Molar mass of NaOH = (1 × Na) + (1 × O) + (1 × H)
= (1 × 22.99) + (1 × 16.00) + (1 × 1.01)
= 39.99 g/mol

Step 2: Calculate the moles of NaOH produced.
The balanced equation tells us that 2 moles of Na react to produce 2 moles of NaOH.
Thus, the ratio of Na to NaOH is 1:1. Therefore, the moles of NaOH produced will be equal to the moles of Na used.

Moles of Na = mass (g) / molar mass (g/mol)
= 20.0 g / 22.99 g/mol
≈ 0.871 mol

Moles of NaOH produced = Moles of Na = 0.871 mol

Step 3: Calculate the mass of NaOH produced.
Mass (g) = Moles × Molar mass

Mass of NaOH = Moles of NaOH × Molar mass of NaOH
= 0.871 mol × 39.99 g/mol
≈ 34.79 g

Therefore, approximately 34.79 grams of sodium hydroxide (NaOH) will be produced when 20.0 grams of sodium (Na) reacts with excess water.

To find the mass of sodium hydroxide produced, you need to first calculate the number of moles of sodium hydroxide produced and then convert it to grams using the molar mass of sodium hydroxide.

1. Calculate the number of moles of sodium metal:
First, find the molar mass of sodium (Na) from the periodic table. Sodium has a molar mass of 22.99 g/mol.
The given mass of sodium is 20.0 g, so you can calculate the number of moles using the formula: moles = mass / molar mass.
moles of sodium = 20.0 g / 22.99 g/mol = 0.8708 mol (rounded to four decimal places).

2. Use the balanced equation to relate the number of moles of sodium to sodium hydroxide:
According to the balanced equation: 2 Na(s) + 2 H2O(l) = 2 NaOH(aq) + H2(g),
The stoichiometric ratio is 2:2, meaning 2 moles of sodium reacts with 2 moles of sodium hydroxide.
Since the ratio is 1:1, the same number of moles of sodium hydroxide is produced.

Therefore, the number of moles of sodium hydroxide is also 0.8708 mol.

3. Calculate the mass of sodium hydroxide:
The molar mass of sodium hydroxide (NaOH) is:
Sodium (Na) = 22.99 g/mol
Oxygen (O) = 15.999 g/mol
Hydrogen (H) = 1.008 g/mol

Molar mass of NaOH = Na + O + H = 22.99 g/mol + 15.999 g/mol + 1.008 g/mol = 39.997 g/mol.
Rounded to two decimal places, the molar mass of sodium hydroxide is approximately 40.00 g/mol.

Now you can calculate the mass of sodium hydroxide produced using the formula: mass = moles x molar mass.
mass of sodium hydroxide = 0.8708 mol x 40.00 g/mol = 34.83 g (rounded to two decimal places).

Therefore, approximately 34.83 grams of sodium hydroxide is produced.