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?
Answer in units of g
To find the mass of O2 produced, we'll need to use the balanced equation for the reaction:
2Na2O2(s) + 2H2O(ℓ) → 4NaOH(aq) + O2(g)
From the equation, we can see that 2 moles of Na2O2 react to produce 1 mole of O2.
Step 1: Calculate the molar mass of Na2O2
The molar mass of Na2O2 is calculated by adding the atomic masses of sodium (Na) and oxygen (O).
Molar mass of Na2O2 = (2 × atomic mass of Na) + atomic mass of O
= (2 × 22.99 g/mol) + 16.00 g/mol
= 45.98 g/mol + 16.00 g/mol
= 61.98 g/mol
Step 2: Calculate the number of moles of Na2O2 reacting
To find the number of moles of Na2O2, divide the given mass by the molar mass.
Number of moles of Na2O2 = Given mass / Molar mass
= 81 g / 61.98 g/mol
≈ 1.31 mol
Step 3: Calculate the number of moles of O2
Since 2 moles of Na2O2 react to produce 1 mole of O2, the number of moles of O2 can be calculated by multiplying the number of moles of Na2O2 by the mole ratio.
Number of moles of O2 = Number of moles of Na2O2 × (1 mole O2 / 2 moles Na2O2)
= 1.31 mol × (1/2)
≈ 0.655 mol
Step 4: Calculate the mass of O2 produced
The mass of O2 can be calculated by multiplying the number of moles of O2 by the molar mass of O2.
Mass of O2 = Number of moles of O2 × Molar mass of O2
= 0.655 mol × 32.00 g/mol
= 20.96 g
Therefore, when 81 g of Na2O2 reacts, approximately 20.96 g of O2 will be produced.
To find the mass of O2 produced when 81 g of Na2O2 reacts, we can follow these steps:
Step 1: Determine the molar mass of Na2O2 and O2.
The molar mass of Na2O2 = 2(22.99 g/mol of Na) + 16.00 g/mol of O = 79.99 g/mol.
The molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol.
Step 2: Calculate the number of moles of Na2O2.
To do this, divide the given mass of Na2O2 by its molar mass:
Number of moles = Mass / Molar mass
Number of moles of Na2O2 = 81 g / 79.99 g/mol ≈ 1.01 mol.
Step 3: Use the mole ratio from the balanced equation to determine the number of moles of O2 produced.
From the balanced equation, the ratio between Na2O2 and O2 is 1:1.
Therefore, the number of moles of O2 produced is also 1.01 mol.
Step 4: Convert the number of moles of O2 to grams.
To do this, multiply the number of moles by the molar mass of O2:
Mass of O2 = Number of moles × Molar mass
Mass of O2 = 1.01 mol × 32.00 g/mol ≈ 32.32 g.
Therefore, approximately 32.32 grams of O2 are produced when 81 grams of Na2O2 reacts.