The balanced reaction equation for the reaction between CH₃OH and O₂ is
2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles 12 24
Reacted moles 12 18
Final moles - 6 12 24
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
= 12 mol x 3 / 2
= 18 mol
All of CH₃OH moles react with O₂.
Hence, the limiting agent is CH₃OH.
Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
Answer:
525.1 g of BaSO₄ are produced.
Explanation:
The reaction of precipitation is:
Na₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) ↓ + 2NaCl (aq)
Ratio is 1:1. So 1 mol of sodium sulfate can make precipitate 1 mol of barium sulfate.
The excersise determines that the excess is the BaCl₂.
After the reaction goes complete and, at 100 % yield reaction, 2.25 moles of BaSO₄ are produced.
We convert the moles to mass: 2.25 mol . 233.38 g/mol = 525.1 g
The precipitation's equilibrium is:
SO₄⁻² (aq) + Ba²⁺ (aq) ⇄ BaSO₄ (s) ↓ Kps
Rust is iron oxide, the corrosion product of iron when exposed to the oxygen in the air. Tin is not iron, so you cannot produce iron oxide from the corrosion of tin. Because the layer of tin on the surface of the steel prevents atmospheric oxygen and moisture from contacting the steel.
Look at the liter man it’s a great way to learn how much every liquid measurement is
