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Willchemistry
2 CH3OH + 3 O2 = 2 CO2 + 4 H2O

2 moles CH3OH ---------------- 3 moles O2
1.33 moles CH3OH ----------- ( moles O2)

moles O2 = 1.33 x 3 / 2

moles O2 = 3.99 / 2

= 1.995 moles of O2

Answer : The number of moles of oxygen needed to burn are, 1.995 moles

Solution : Given,

Moles of methyl alcohol, [tex]CH_3OH[/tex] = 1.33 mole

The given balanced chemical reaction is,

[tex]2CH_3OH+3O_2\rightarrow 2CO_2+4H_2O[/tex]

From the balanced reaction, we conclude that

As, 2 moles of methyl alcohol react with 3 moles of oxygen

So, 1.33 moles of methyl alcohol react with [tex]\frac{3}{2}\times 1.33=1.995[/tex] moles of oxygen

Therefore, the number of moles of oxygen needed to burn are, 1.995 moles

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