Answer:
101 L
Explanation:
35.0 g KOH ÷ 56.09 g/mol KOH × (1 mol H2O/ 1 mol KOH) × 18 g/mol H2O = 11.2 g H2O
35.0 g HCl ÷ 36.45 g/mol HCl × (1 mol H2O/ 1 mol HCl) × 18 g/mol H2O = 17.3 g H2O
35.0 g KOH is the limiting reactant
2NaClO₃ → 2NaCl + 3O₂
mole ratio of NaClO₃ to O₂ is 2 : 3
∴ if moles of NaClO₃ = 12 mol
then moles of O₂ =
= 18 mol
Mass of O₂ = mol of O₂ × molar mass of O₂
= 18 mol × 16 g/mol
= 288 g
So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction. If you should have used another reaction then just follow the same steps and you'll get your answer.