Question

Methanol (CH3OH) is the fuel used by racecars in the Indianapolis 500. It burns in the engine according to the equation below.

How many moles of oxygen gas are necessary to react completely with 425 moles of methanol?
2CH3OH() + 302(g) → 2CO2(g) + 4H2O(g)

Answer 1

637.5

Step-by-step explanation:

The reaction is balanced so using the coefficients you have that 2 moles of methanol reacts with 3 moles oxygen.

moles of oxygen = 3/2 moles of methanol

moles of oxygen = 3/2 *425 = 637.5 moles

Answer 2

637.5 moles of oxygen gas are necessary to react completely with 425 moles of methanol.

Step-by-step explanation:

You know the following equation:

2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)

Observing the stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you can see that to react with 2 moles of methanol, you need 3 moles of oxygen gas (O₂).

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

Then:

`x=(c*b)/(a)`

Then, the rule of three is applied to calculate the amount of moles of oxygen gas to react completely with 425 moles of methanol, knowing the stoichiometry of the reaction. This rule applies as follows: if by stoichiometry 2 moles of methane must react with 3 moles of oxygen gas, when 425 moles of methanol react how many moles of oxygen gas are needed?

`moles of oxygen gas=(425moles of methanol*3moles of oxygen)/(2 moles of methanol) \\`

moles of oxygen gas=637.5

Then, 637.5 moles of oxygen gas are necessary to react completely with 425 moles of methanol.