If 120 ml of oxygen is collected over water at 27 degrees celsius & 740 mm pressure what will the volume of the dry gas be at stp

                       V₂ = 106.33 ml

Explanation:

Data Given:

                 Initial Temperature  =  T₁  =  27 °C + 273.15 = 300.15 K

                 Initial Pressure  =  P₁  =  740 mmHg

                 Initial Volume  =  V₁  =  120 ml

                 Final Temperature  =  T₂  =  0 °C + 273.15 = 273.15 K

                 Final Pressure  =  P₂  =  760 mmHg

                 Fnal Volume  =  V₂  =  ?

Formula Used:

Let's assume that the Oxygen gas is acting as an Ideal gas, the according to Ideal Gas Equation,

                       P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                       V₂ = P₁ V₁ T₂/ T₁ P₂

Putting values,

                       V₂ = 740 mmHg × 120 ml × 273.15 K / 300.15 K × 760 mmHg

                       V₂ = 106.33 ml

not clear

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"