At 25°c, the standard enthalpy of combustion of gaseous propane (c3h8) is –2219.0 kj per mole of propane, and the standard enthalpy of combustion of gaseous propylene (c3h6) is –2058.3 kj per mole of propylene.
what is the standard enthalpy change for the following reaction at 25°c? c3h6(g) + h2(g) → c3h8(g)substance∆h°f (kj/mol)co2(g)–393.5h2o(l)–285.8

The standard enthalpy change for the given reaction is 868.05 kJ

Explanation:

The chemical equation for the combustion of propane follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

The enthalpy of formation of is -140.7 kJ/mol

The chemical equation for the combustion of propane follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

The enthalpy of formation of is -1008.75 kJ/mol

The given chemical equation follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

Hence, the standard enthalpy change for the given reaction is 868.05 kJ

"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"