Calculate the percentage of water in the compound CaSO4 x 6H2O

The percentage of water in the compound CaSO₄.6H₂O is 44.26%

Since MW CaSO₄. 6H₂O :

= 40 + 32 + 4.16 + 12.1 + 6.16

= 244

Now

MW 6H2O :

= 12.1 + 6.16

= 108

So, percentage of water = (108 : 244) x 100%

= 44.26

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The percentage of water in the compound CaSO₄.6H₂O = 44.26%

Given

Compound CaSO₄. 6H₂O

Required

The percentage of water

Solution

MW CaSO₄. 6H₂O :

= 40 + 32 + 4.16 + 12.1 + 6.16

= 244

MW 6H2O :

= 12.1 + 6.16

= 108

%water = (108 : 244) x 100%

%water = 44.26

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :