Question

Consider the combustion of methanol at some high temperature in a constant-pressure reaction chamber: 2ch3oh (g) + 3o2 (g) \longrightarrow ⟶ 2co2 (g) + 4h2o (g) if you react 10 l of oxygen gas with 18 l of methanol gas, what will the final volume be?

Answer 1

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 2

Answer : The final volume of the reaction mixture is 31.33 L

Explanation :

The given balanced chemical equation is ,

$2CH_{3}OH (g) + 3O_{2} (g) \rightarrow 2CO_{2} (g) + 4H_{2}O (g)$

Step 1 : Find limiting reactant

In order to find the limiting reactant, we will divide the given volumes by the stoichiometric coefficient from the balanced equation. The reactant that gives lower volume is the limiting reactant.

$\frac{10 L O_{2} }{3} = 3.33 L O_{2}$

$\frac{18 L CH_{3}OH }{2} = 9 L CH_{3}OH$

Since the volume of O₂ is lower, it is the limiting reactant.

Step 2 : Using limiting reactant, find out the volumes of product formed.

The amount of CO₂ formed can be calculated as,

$10 L O_{2} \times \frac{2 CO_{2}}{3 O_{2}} = 6.67 L CO_{2}$

The amount of H₂O formed can be calculated as,

$10 L O_{2} \times \frac{4 H_{2}O}{3 O_{2}} = 13.33 L H_{2}O$

Total volume of products = 6.67 L + 13.33 L = 20 L

Step 3 : Find the volume of excess reactant.

The amount of methanol that reacts with the given amount of O₂ is

$10 L O_{2} \times \frac{2 CH_{3}OH}{3 O_{2}} = 6.67 L CH_{3}OH$

But the initial amount of methanol is 18 L.

The amount of excess reactant is 18 L - 6.67 L = 11.33 L

The total volume of gases after the reactant is complete = total volume of products + excess reactant = 20 L + 11.33 L = 31.33 L

The final volume of the reaction mixture is 31.33 L