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 24 g of
Na2O2 react?
What mass of water is needed to react completely
with the Na2O2?
#1. Balance the chemical reaction first.
2 Na2O2 + 2 H2O −--> 4 NaOH + O2
#2. Then calculate the molar mass of Na2O2 and O2. From the periodic table, Na - 23 and O - 16. Thus,
Na2O2 : 2*23 + 2*16 = 78 g/mol
O2 : 2*16 = 32 g/mol
#3. Get the moles of Na2O2 from the given 24 g Na2O2, by dividing this mass by the molar mass:
24 g / 78 g/mol = 0.30769 mol Na2O2
#4. From the balanced chemical reaction, we make a mole ratio of O2 : Na2O2. That is, for every mole of O2 produced, there are 2 moles of Na2O2 reacted, or
1 mol O2 / 2 mol Na2O2
#5. Multiply the moles of Na2O2 by the ratio to cancel units:
0.30769 mol Na2O2 * 1 mol O2 / 2 mol Na2O2 = 0.15385 mol O2
#6. Finally, multiply the moles of O2 to its molar mass to get the mass:
0.15385 mol O2 * 32 g/mol = 4.923 g O2
Now, try doing the second question.
Hope this helps~ `u`
Well, if you're looking for some chemical comedy, you've come to the right place! Let's calculate the mass of O2 and the quantity of water needed.
To find the mass of O2 produced, we need to balance the equation first. The balanced equation for this reaction is:
2 Na2O2(s) + 2 H2O(ℓ) → 4 NaOH(aq) + O2(g)
From here, we can see that for every 2 moles of Na2O2 reacted, we get 1 mole of O2. We'll follow a step-by-step approach to reach the answer.
Step 1: Convert the mass of Na2O2 to moles.
Molar mass of Na2O2 = 22.99 x 2 + 16.00 x 2 = 111.99 g/mol
Moles of Na2O2 = Mass of Na2O2 / Molar mass of Na2O2 = 24 g / 111.99 g/mol
Step 2: Use the mole ratio to find moles of O2 produced.
According to the balanced equation, 2 moles of Na2O2 react to produce 1 mole of O2. So, we can write:
Moles of O2 = Moles of Na2O2 / 2
Step 3: Convert moles of O2 to mass.
Molar mass of O2 = 16.00 g/mol
Mass of O2 = Moles of O2 x Molar mass of O2
Follow these steps, and you'll end up with the answer – the mass of O2 produced!
As for the mass of water needed to react completely, we can use the stoichiometry of the balanced equation. The coefficient of H2O in the equation is 2, which means for every 2 moles of Na2O2, we require 2 moles of H2O. You can then use the molar mass of H2O and follow the same procedure outlined above to find the mass of water needed.
And remember, if chemistry gets too overwhelming, just add a little "haha" and it will all become pun-tastic!
To find the mass of O2 produced when 24 g of Na2O2 react, we need to balance the equation first.
The balanced equation for the reaction is:
2Na2O2(s) + 2H2O(ℓ) --> 4NaOH(aq) + O2(g)
From the balanced equation, we can see that 2 moles of Na2O2 react to produce 1 mole of O2.
First, let's calculate the molar mass of Na2O2:
The molar mass of Na is 22.99 g/mol.
The molar mass of O is 16.00 g/mol.
The molar mass of Na2O2 = (2 * molar mass of Na) + (2 * molar mass of O)
= (2 * 22.99 g/mol) + (2 * 16.00 g/mol)
= 45.98 g/mol + 32.00 g/mol
= 77.98 g/mol
Now, we can calculate the moles of Na2O2 by dividing the given mass by the molar mass:
Moles of Na2O2 = Mass / Molar mass
= 24 g / 77.98 g/mol
= 0.3077 mol
Since we know that 2 moles of Na2O2 produce 1 mole of O2, we can calculate the moles of O2 produced:
Moles of O2 = Moles of Na2O2 / 2
= 0.3077 mol / 2
= 0.1538 mol
Finally, we can calculate the mass of O2 produced by multiplying the moles of O2 by its molar mass:
Mass of O2 = Moles of O2 * Molar mass
= 0.1538 mol * 32.00 g/mol
= 4.922 g
Therefore, 4.922 g of O2 is produced when 24 g of Na2O2 react.
Now, let's find the mass of water needed to react completely with the Na2O2.
From the balanced equation, we can see that 2 moles of Na2O2 react with 2 moles of H2O.
To calculate the mass of water, we need to find the moles of Na2O2 using the given mass and molar mass:
Moles of Na2O2 = Mass of Na2O2 / Molar mass of Na2O2
= 24 g / 77.98 g/mol
= 0.3077 mol
Since 2 moles of Na2O2 react with 2 moles of H2O, the moles of water needed will be the same.
Therefore, the mass of water needed to react completely with the Na2O2 is:
Mass of water = Moles of H2O * Molar mass of H2O
= 0.3077 mol * 18.02 g/mol
= 5.541 g
Hence, 5.541 g of water is needed to react completely with the Na2O2.
To find the mass of O2 produced when 24 g of Na2O2 reacts, we need to balance the chemical equation first:
2 Na2O2(s) + 2 H2O(ℓ) → 4 NaOH(aq) + O2(g)
From the balanced equation, we can see that 2 moles of Na2O2 will produce 1 mole of O2.
First, we need to calculate the number of moles of Na2O2 in 24 g:
Molar mass of Na2O2 = (2 * atomic mass of Na) + (2 * atomic mass of O)
= (2 * 22.99 g/mol) + (2 * 16.00 g/mol)
= 45.98 g/mol + 32.00 g/mol
= 77.98 g/mol
Number of moles of Na2O2 = mass / molar mass
= 24 g / 77.98 g/mol
≈ 0.308 moles
Since 2 moles of Na2O2 produce 1 mole of O2, the number of moles of O2 produced is also 0.308.
Next, we can find the mass of O2 produced using the molar mass of O2:
Molar mass of O2 = 2 * atomic mass of O
= 2 * 16.00 g/mol
= 32.00 g/mol
Mass of O2 produced = number of moles of O2 * molar mass of O2
= 0.308 moles * 32.00 g/mol
= 9.856 g
Therefore, when 24 g of Na2O2 react, approximately 9.856 g of O2 is produced.
To find the mass of water needed to react completely with the Na2O2, we use the stoichiometry of the balanced equation.
From the balanced equation:
2 Na2O2(s) + 2 H2O(ℓ) → 4 NaOH(aq) + O2(g)
We can see that 2 moles of Na2O2 react with 2 moles of H2O to produce 4 moles of NaOH.
Since we know the moles of Na2O2 (0.308 moles), we can use the stoichiometry to find the moles of H2O needed:
Moles of H2O = 0.308 moles * (2 moles H2O / 2 moles Na2O2)
= 0.308 moles
Finally, we can find the mass of water needed using the molar mass of water:
Molar mass of H2O = (2 * atomic mass of H) + atomic mass of O
= (2 * 1.01 g/mol) + 16.00 g/mol
= 2.02 g/mol + 16.00 g/mol
= 18.02 g/mol
Mass of water needed = moles of H2O * molar mass of H2O
= 0.308 moles * 18.02 g/mol
≈ 5.54 g
Therefore, approximately 5.54 g of water is needed to react completely with 24 g of Na2O2.