Final answer:
To calculate the mass of sodium hydroxide that would contain 22.3g of oxygen, use the balanced chemical equation and the molar mass of sodium hydroxide. The mass of sodium hydroxide containing 22.3 grams of oxygen is 356.8 grams.
Step-by-step explanation:
To calculate the mass of sodium hydroxide that would contain 22.3g of oxygen, we need to use the balanced chemical equation and the molar mass of sodium hydroxide. From the balanced equation: 4NaOH(aq) + 2S(s) + 302(g) -> 2Na2SO4(aq) + 2H₂O(l), we can see that 2 moles of sodium hydroxide react with 1 mole of oxygen to form 2 moles of water. This means that 1 mole of sodium hydroxide reacts with 0.5 moles of oxygen to form 1 mole of water. Now, we can use the molar mass of sodium hydroxide (40.00 g/mol) to convert the moles of oxygen to grams of sodium hydroxide.
1 mole of sodium hydroxide = 40.00 grams of sodium hydroxide
0.5 moles of oxygen = 22.3 grams of oxygen
To find the mass of sodium hydroxide, we use the equation:
(22.3 grams of oxygen) x (40.00 grams of sodium hydroxide / 0.5 moles of oxygen) = 356.8 grams of sodium hydroxide containing 22.3 grams of oxygen.