What mass, in grams, of sodium hydroxide (NaOH) is produced if 20 grams of sodium metal(Na) reacts with excess water?
1. Write the equation and balance it.
2. Convert 20 g Na metal to moles. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Na metal to mols NaOH.
4. Now convert moles NaOH to grams. g = moles x molar mass.
To determine the mass of sodium hydroxide (NaOH) produced when 20 grams of sodium metal (Na) reacts with excess water, we need to understand the balanced chemical equation for the reaction between sodium and water.
The balanced chemical equation for this reaction is:
2 Na + 2 H2O → 2 NaOH + H2
From this equation, we can see that for every 2 moles of sodium (Na), we obtain 2 moles of sodium hydroxide (NaOH).
Now, let's calculate the molar mass of sodium hydroxide (NaOH):
- Molar mass of sodium (Na) = 22.99 g/mol
- Molar mass of oxygen (O) = 16 g/mol
- Molar mass of hydrogen (H) = 1.01 g/mol
Adding up the molar masses:
- Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 39.00 g/mol
Next, we determine the number of moles of sodium (Na) in 20 grams:
- Moles of Na = mass ÷ molar mass = 20 g ÷ 22.99 g/mol = 0.870 moles
Since the molar ratio between Na and NaOH is 1:1, the number of moles of sodium hydroxide (NaOH) produced will also be 0.870 moles.
Finally, we can calculate the mass of sodium hydroxide (NaOH) using its molar mass:
- Mass of NaOH = moles × molar mass = 0.870 moles × 39.00 g/mol = 33.93 grams
Therefore, 33.93 grams of sodium hydroxide (NaOH) will be produced when 20 grams of sodium metal (Na) reacts with excess water.
To determine the mass of sodium hydroxide produced, we first need to write and balance the chemical equation for the reaction:
2Na + 2H2O → 2NaOH + H2
From the balanced equation, we can see that 2 moles of sodium (Na) react with 2 moles of water (H2O) to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of hydrogen gas (H2).
Now, we need to calculate the amount of sodium hydroxide produced. Given that 20 grams of sodium (Na) is reacted, we need to convert the mass of sodium to moles using the molar mass of sodium (Na):
Molar mass of Na = 22.99 g/mol
Number of moles of Na = Mass of Na / Molar mass of Na
= 20 g / 22.99 g/mol
= 0.869 moles
Since 2 moles of sodium hydroxide (NaOH) are produced for every 2 moles of sodium (Na), we can conclude that 0.869 moles of sodium (Na) would produce 0.869 moles of sodium hydroxide (NaOH).
Finally, we can convert moles of sodium hydroxide to grams using the molar mass of sodium hydroxide (NaOH):
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
= 39.00 g/mol
Mass of sodium hydroxide = Number of moles of NaOH * Molar mass of NaOH
= 0.869 * 39.00 g/mol
= 33.891 grams
Therefore, approximately 33.891 grams of sodium hydroxide (NaOH) is produced when 20 grams of sodium metal (Na) reacts with excess water.
2Na + S --> Na2S
20g Na x (1 mole Na)/(23g Na) = 0.87 mol Na
0.87 mole Na x (1 mol Na2S)/(2 mol Na) x (55.1g Na2S)/(1 mol Na2S)
=23.97g Na2S