Question

Methanol (ch3oh) can react with oxygen gas to produce formaldehyde (h2co) and water. How many mole of formaldehyde can be produced by reacting 4.0 moles of methanol with 4.0 moles of oxygen gas

Answer 1

Answer : The number of moles of formaldehyde produced will be, 4 moles.

Explanation :

Moles of $CH_3OH$ = 4.0 mole

Moles of $O_2$ = 4.0 mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be :

$2CH_3OH(l)+O_2(g)\rightarrow 2H_2CO(l)+2H_2O(l)$

From the balanced reaction we conclude that

As, 2 moles of $CH_3OH$ react with 1 mole of $O_2$

So, 4 moles of $CH_3OH$ react with $\frac{4}{2}=2$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $CH_3OH$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2CO$.

As, 2 moles of $CH_3OH$ react to give 2 moles of $H_2CO$

So, 4 moles of $CH_3OH$ react to give 4 moles of $H_2CO$

Therefore, the number of moles of formaldehyde produced will be, 4 moles.

Answer 2

The number of moles of formaldehyde can be produced by reacting 4.0 moles of methanol with 4.0 moles of oxygen gas is 8 moles

explanation

write   a balanced chemical reaction

that is 2 CH3OH (l) +  O2 (g) →  2HCOH (l)+ 2H2O (l)

from the reaction above  2 moles of CH3OH reacted with 1 moles of O2 to form 2 moles of HCOH and two moles of H2O

This imply that O2 is the limiting reagent ,therefore by use of mole ratio between  O2 : HCOH which is 1 :2 the moles of HCOH= 2 x4 =  8 moles