Sodium peroxide reacts vigorously with water to produce sodium hydroxide and oxygen. The unbalanced equation is
Na2O2(s) + H2O(ℓ)→ NaOH(aq) + O2(g). What mass of O2 is produced when 81 g of Na2O2
react?
First, we balance the chemical reaction:
2 Na2O2 + 2 H2O → 4 NaOH + O2
Then, we need to determine the molar masses of Na2O2 and O2 in order to convert them to moles. To get them, we need the mass of each element in the formula and add. In the periodic table, Na = 23 and O = 16. Thus the respective molar masses are:
Na2O2: 2*23 + 2*16 = 78 g/mol
O2: 2*16 = 32 g/mol
Then we use the given, 81 g of Na2O2. To get the moles of Na2O2, we divide the mass by the molar mass:
81 g / (78 g/mol) = 1.03846 moles Na2O2
From the balanced reaction, 1 mole of O2 is being produced for every 2 moles of Na2O2 reacted. Thus,
1.03846 moles Na2O2 * (1 mole O2 / 2 moles Na2O2) = 0.5192 moles O2
Finally, we multiply this by the molar mass of O2 to get its mass:
0.5192 * 32 = 16.62 g O2
Hope this helps~ `u`
Thank YOU!!!!
To determine the mass of O2 produced when 81 g of Na2O2 reacts, we need to first balance the equation:
2Na2O2(s) + 2H2O(ℓ) → 4NaOH(aq) + O2(g)
The balanced equation shows that for every 2 moles of Na2O2, we produce 1 mole of O2.
Next, we'll calculate the molar mass of Na2O2 and O2.
The molar mass of Na2O2 can be calculated as follows:
2(Na) + 2(O) = (2 * 22.99 g/mol) + (2 * 16.00 g/mol) = 61.98 g/mol
The molar mass of O2 is:
2(O) = 2 * 16.00 g/mol = 32.00 g/mol
Now, we'll use the molar mass of Na2O2 to calculate the number of moles of Na2O2 in 81 g:
moles of Na2O2 = mass (g) / molar mass (g/mol)
moles of Na2O2 = 81 g / 61.98 g/mol = 1.31 mol
Since the balanced equation shows that 2 moles of Na2O2 produce 1 mole of O2, we can determine the number of moles of O2 produced:
moles of O2 = (1.31 mol Na2O2) / 2 = 0.655 mol
Finally, we'll calculate the mass of O2 produced using the moles of O2 and its molar mass:
mass of O2 = moles of O2 * molar mass of O2
mass of O2 = 0.655 mol * 32.00 g/mol = 20.96 g
Therefore, when 81 g of Na2O2 reacts, 20.96 g of O2 is produced.
To determine the mass of O2 produced when 81 g of Na2O2 reacts, we need to use stoichiometry to find the mole ratio between Na2O2 and O2, and then convert the moles of O2 to grams.
Step 1: Write the balanced equation:
2 Na2O2(s) + 2 H2O(ℓ) → 4 NaOH(aq) + O2(g)
Step 2: Calculate the molar mass of Na2O2 and O2:
Na2O2: Sodium (Na) has a molar mass of 22.99 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Since there are two sodium atoms and two oxygen atoms in Na2O2, the molar mass of Na2O2 is:
2(22.99 g/mol) + 2(16.00 g/mol) = 87.98 g/mol
O2: Oxygen (O) has a molar mass of 16.00 g/mol.
Step 3: Calculate the number of moles of Na2O2:
Moles of Na2O2 = Mass of Na2O2 / Molar mass of Na2O2
Moles of Na2O2 = 81 g / 87.98 g/mol ≈ 0.920 mol
Step 4: Use the mole ratio from the balanced equation to find the moles of O2 produced:
From the balanced equation, we see that 2 moles of Na2O2 produces 1 mole of O2.
Moles of O2 produced = Moles of Na2O2 × (1 mole O2 / 2 moles Na2O2)
Moles of O2 produced = 0.920 mol × (1 mol / 2 mol) = 0.460 mol
Step 5: Convert moles of O2 to grams:
Mass of O2 = Moles of O2 × Molar mass of O2
Mass of O2 = 0.460 mol × 16.00 g/mol = 7.36 g
Therefore, when 81 g of Na2O2 reacts, 7.36 g of O2 is produced.