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.

The answer for the following problem has been mentioned below.

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 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 = × 0.0821 × 425

                   11.25 =   × 34.8925

                  180 = m × 34.8925

                  m =

                  m = 5.158 grams

Therefore the mass of the of oxygen is 5.158 grams

Now;

As we know;

where;

represents the mass of the oxygen

represents the gram molecular mass of the oxygen

represents the mass of the water

represents the gram molecular mass of water

    From the above given formula,

      =

where;

Gram molecular weight of water = 18.0 u

    = 5.802 grams

Therefore the mass of the water is 5.802 grams.

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current