Question

How many molecules of carbon monoxide gas can be produced from 395 grams of oxygen gas ?

Answer 1

The answer for the following problem is mentioned below.

• Therefore 148.125 × 10^-23 molecules of the cabon monoxide gas is produced.

Explanation:

Given:

mass of the oxygen gas = 395 grams

We know;

For the production of the carbon monoxide;

The equation is,

Before balancing the equation;

C + $O_{2}$ → CO

After balancing the equation;

2 C +$O_{2}$ → 2 CO

where,

C = carbon molecule

O = oxygen molecule

CO = carbon monoxide

For the equation,

  32 grams of oxygen gas   →  2 × 6.023 × 10^-23 molecules

        395 grams of oxygen gas   →    ?

     

      = $\frac{2*6.023*10^{-23}* 395}{32}$

   = 148.125 × 10^-23 molecules of the carbon monoxide

Therefore 148.125 × 10^-23 molecules of the cabon monoxide gas is produced.