what mass in grams of sodium hydroxide is produced if 20.0g of sodium metal reacts with excess water according ro the chemical equation, Na(s)+H2O4->naoh(aq)+h2(g)
mols Na metal = grams/molar mass = ?
Convert mols Na to mols NaOH using the coefficients in the balanced equation.
Then convert mols NaOH to g. g = mols x molar mass.
To find the mass of sodium hydroxide produced, we need to determine the balanced equation and use stoichiometry.
The balanced equation for the reaction is:
2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)
Now, let's calculate the molar mass of sodium hydroxide (NaOH):
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Next, let's determine the number of moles of sodium metal (Na) using its molar mass:
Molar mass of Na = 22.99 g/mol
Number of moles of Na = Mass of Na / Molar mass of Na
= 20.0 g / 22.99 g/mol
≈ 0.8708 mol
From the balanced equation, we can see that 2 moles of NaOH are produced for every 2 moles of Na. This means the mole ratio of Na to NaOH is 1:1.
Therefore, the number of moles of sodium hydroxide (NaOH) produced is also 0.8708 mol.
Finally, let's calculate the mass of sodium hydroxide produced:
Mass of NaOH = Number of moles of NaOH * Molar mass of NaOH
= 0.8708 mol * 40.00 g/mol
= 34.832 g
Therefore, the mass of sodium hydroxide produced is approximately 34.832 grams.