2CH3OH + 3O2-> 2CO2 + 4 H2O
wHAT IS THE AMOUNT OF OXYGEN IS REQUIRED TO COMPLETRLY BURN 5 MOL OF METHANOL?
To determine the amount of oxygen required to completely burn 5 mol of methanol (CH3OH), we will use the balanced chemical equation provided:
2CH3OH + 3O2 -> 2CO2 + 4H2O
From the equation, we can see that 3 moles of oxygen are required to burn 2 moles of methanol.
1. Calculate the amount of methanol required to burn 5 mol of methanol using the ratio from the balanced equation:
(5 mol CH3OH) / (2 mol CH3OH) = 2.5 mol CH3OH
2. Use the ratio from the balanced equation to determine the required amount of oxygen:
(2.5 mol CH3OH) * (3 mol O2 / 2 mol CH3OH) = 3.75 mol O2
Therefore, the amount of oxygen required to completely burn 5 mol of methanol is 3.75 mol.
To determine the amount of oxygen required to completely burn 5 moles of methanol (CH3OH) using the balanced chemical equation:
2CH3OH + 3O2 -> 2CO2 + 4H2O
We will use stoichiometry.
Stoichiometry is a method of using the balanced chemical equation to calculate the quantities of substances involved in a chemical reaction.
From the balanced equation, we can see that the ratio of methanol to oxygen is 2:3. This means that for every 2 moles of methanol, we need 3 moles of oxygen to react completely.
So for 5 moles of methanol, we can set up a proportion to find the amount of oxygen required:
(5 moles CH3OH) / (2 moles CH3OH) = (x moles O2) / (3 moles O2)
Cross-multiplying and solving for x, we get:
5 moles CH3OH × (3 moles O2 / 2 moles CH3OH) = 7.5 moles O2
Therefore, to completely burn 5 moles of methanol, we need 7.5 moles of oxygen.
5 mol CH3OH x (3 mols O2/2 mol CH3OH) = 5*3/2 = ? mols O2.
grams O2, if you need it = mols O2 x molar mass O2.