Students massed 50g of sodium (2Na) and 27g of water (2H2O). They combined the 2 reactants and produced the products.
2Na + 2H2O --> 2NaOH + H2
If 42g of sodium hydroxide (2NaOH) was produced, what is the mass of the hydrogen gas produced (H2)?
To find the mass of hydrogen gas produced, we need to calculate the difference in mass before and after the reaction.
Mass of sodium hydroxide produced (2NaOH) = 42g
Since the balanced equation shows that the ratio of sodium hydroxide (2NaOH) to hydrogen gas (H2) is 2:1, we can determine the mass of hydrogen produced.
Mass of hydrogen gas produced (H2) = Mass of sodium hydroxide produced (2NaOH) / 2
Mass of hydrogen gas produced (H2) = 42g / 2
Mass of hydrogen gas produced (H2) = 21g
Therefore, the mass of hydrogen gas produced is 21 grams.
To find the mass of hydrogen gas produced, we need to determine the amount of moles of sodium hydroxide (2NaOH) produced and then use the stoichiometry of the balanced equation to find the amount of moles of hydrogen gas (H2) produced. Finally, we can convert the moles of hydrogen gas to grams using the molar mass of hydrogen.
1. Calculate the amount of moles of sodium hydroxide produced:
Mass of sodium hydroxide = 42g
Molar mass of sodium hydroxide (2NaOH) = 2(22.99 g/mol) + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol
Amount of moles of sodium hydroxide produced = mass of sodium hydroxide / molar mass of sodium hydroxide
= 42g / 39.00 g/mol
ā 1.08 mol
2. Use the stoichiometry of the balanced equation to find the amount of moles of hydrogen gas produced:
According to the balanced equation, 2 moles of sodium (2Na) react with 2 moles of water (2H2O) to produce 2 moles of sodium hydroxide (2NaOH) and 1 mole of hydrogen gas (H2).
So, the amount of moles of hydrogen gas produced = 1.08 mol
3. Convert the amount of moles of hydrogen gas to grams using the molar mass of hydrogen:
Molar mass of hydrogen (H2) = 2(1.01 g/mol) = 2.02 g/mol
Mass of hydrogen gas produced = amount of moles of hydrogen gas * molar mass of hydrogen
= 1.08 mol * 2.02 g/mol
ā 2.18 g
Therefore, the mass of the hydrogen gas produced (H2) is approximately 2.18 grams.
To find the mass of the hydrogen gas produced (H2), we first need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
The molar mass of sodium (Na) is 23 g/mol, and the molar mass of water (H2O) is 18 g/mol. Let's calculate the number of moles for each reactant:
Moles of sodium (Na) = (mass of sodium / molar mass of sodium) = (50 g / 23 g/mol) = 2.17 mol
Moles of water (H2O) = (mass of water / molar mass of water) = (27 g / 18 g/mol) = 1.5 mol
According to the balanced equation, 2 moles of sodium hydroxide (NaOH) are produced for every 2 moles of sodium (Na) and 2 moles of water (H2O) reacted.
Moles of sodium hydroxide (NaOH) = 2 moles
Now, let's determine the moles of hydrogen gas (H2) produced by using the stoichiometric ratio:
Moles of hydrogen gas (H2) = (moles of sodium hydroxide (NaOH) / stoichiometric ratio of hydrogen gas (H2)) = (2 mol / 2) = 1 mol
Finally, we can calculate the mass of hydrogen gas (H2) using the molar mass of hydrogen (H):
Mass of hydrogen gas (H2) = (moles of hydrogen gas (H2) x molar mass of hydrogen (H)) = (1 mol x 2 g/mol) = 2 g
Therefore, the mass of the hydrogen gas produced (H2) is 2 grams.