Question

How many grams of water are produced when 4.50 L of

oxygen gas was consumed. The reaction temperature
and pressure are 425 K and 2.50 atm.

Answer 1

The answer for the following problem has been mentioned below.

• Therefore the mass of the water is 5.802 grams.

Step-by-step explanation:

Given:

volume of oxygen (V) = 4.50 L

Temperature (T) = 425 K

pressure of oxygen (P) = 2.50 atm

Gram molecular mass of oxygen (M) = 16.0 grams

To calculate:

mass of water (m)

We know;

According to the ideal gas equation;

P × V = n × R × T

As we know;

no of moles =
`(m)/(M)`

m represents the mass of oxygen (m)

M represents the Gram molecular mass (M)

According to above mentioned equation;

P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

where,

the value of R is 0.0821 L atm/K moles

Substituting the values in the above equation;

2.50 × 4.50 =
`(m)/(16.0)` × 0.0821 × 425

11.25 =
`(m)/(16.0)` × 34.8925

180 = m × 34.8925

m =
`(180)/(34.8925)`

m = 5.158 grams

Therefore the mass of the of oxygen is 5.158 grams

Now;

As we know;

`(m_(1) )/(M_(1) ) = (m_(2) )/(M_(2) )`

where;

`m_(1)` represents the mass of the oxygen

`M_(1)` represents the gram molecular mass of the oxygen

`m_(2)` represents the mass of the water

`M_(2)` represents the gram molecular mass of water

From the above given formula,

`(5.158)/(16.0)` =
`(m_(2) )/(18)`

where;

Gram molecular weight of water = 18.0 u

`m_(2)` = 5.802 grams

Therefore the mass of the water is 5.802 grams.