Question

The reaction between methanol and oxygen gas produces water vapor and carbon dioxide. 2CH3OH(l)+3O2(g)⟶4H2O(g)+2CO2(g) Three sealed flasks contain different amounts of methanol and oxygen. Based on the molecular view of the three flasks, which would produce the largest quantity of total product? The flask contains 3 molecules of C H 3 O H and 3 molecules of O 2. The flask contains 1 molecule of C H 3 O H and 6 molecules of O 2. The flask contains 4 molecules of C H 3 O H and 2 molecule of O 2. For the flask which produces the largest quantity of total product, how many molecules of H2O will be formed? molecules of H2O :

Answer 1

Answer:

The flask that produces the largest quantity of product is  the second flask

The molecules of H 2 O formed is  $X = 8 \ molecules$

Explanation:

From the question we are told that

  The reaction is  

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

 The first flask contains  3 molecules of C H 3 O H and 3 molecules of O 2

 The  second flask contains  1 molecule of C H 3 O H and 6 molecules of O 2

  The third flask contains  4 molecules of C H 3 O H and 2 molecule of O 2.

Looking at the three flasks we can see that base on molecular view the flask which produces the largest quantity of total product is second flask

 From the balanced equation we see that 2 moles of  C H 3 O H  are required to react with 3 moles of  oxygen hence O 2  is the limiting reactant

Generally 1 mole of any  substance is  $1 * N_A$ of that substance

Here $N_A$  is the Avogadro number with the value  $N_A = 6.02214076 * 10^{23} \ molecules$

So  

Looking at the balanced equation we see that 3 moles of O 2 (3 *  $N_A$  molecules of O 2 ) produces 4 moles of  H2O(4 *  

Then 6 molecules of O 2 will produce X molecules of  H 2 O

So  

     $X = \frac{6 * (4 * N_A }{3 * N_A}$

=>   $X = 8 \ molecules$