The balanced reaction is
2CH3OH (g) + 3O2 (g) ⟶ 2CO2 (g) + 4H2O (g)
as per equation two moles of methanol gas will react with 3 moles of oxygen
one mole of gas occupies 22.4 L of volume
so the moles and volume goes in same ratio
it means two unit volume of methanol will react with three unit volume of oxygen
therefore 1L of methanol gas will react with 3 /2 L of oxygen
Or 18 L of methanol gas will react with 3 x 18 /2 = 27 L of oxygen
So here oxygen is limiting reagent
As per balanced equation
10L of oxygen will react with = 2 X 10 /3 L of methanol = 6.67 L of methanol gas to give 6.67 L of CO2 gas and 13.33 L of water gas
So overall there will be = 18 - 6.67 L of left out methanol = 11.33 L
And 6.67 L of CO2 + 13.33 L of water = 20 L
Total volume of gas = 11.33+ 20 = 31.33 L
Answer:
is there choices you have to pick from
Explanation:
or do you have to describe a covalent bond ?
Mg + 2️⃣H20 ➡️ Mg(OH)2 + H2
Answer:
If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to reduce a chemical species while simultaneously forming iodine.
