Answer: 1.
moles
2. 90 mg
Explanation:

According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus
moles of ozone is removed by =
moles of sodium iodide.
Thus
moles of sodium iodide are needed to remove
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide=
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of
.
Answer:
Explanation:
stoichiometry of C₂H₂ to H₂O is 2:2.
Number of moles of C₂H₂ = molar mass of C₂H₂
Since the molar mass of C₂H₂ is 26 g/mol.
Number of C₂H₂ moles reacted = 64.0 g / 26 g/mol = 2.46 mol.
according to a molar ratio of 2:2.
the number of H₂O moles formed = a number of C₂H₂ moles reacted.
Therefore the number of H₂O moles produced = 2.46 mol
I think it’s A but I’m not sure.