how many moles of oxygen are needed to burn 1.33 moles of methyl alcohol (ch3oh)?2ch3oh 3o2 2co2 4h2o
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, \(CH_3OH\) = 1.33 mole
The given balanced chemical reaction is,
\(2CH_3OH+3O_2\rightarrow 2CO_2+4H_2O\)
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 \(\frac{3}{2}\times 1.33=1.995\) moles of oxygen
Therefore, the number of moles of oxygen needed to burn are, 1.995 moles