What mass of oxygen is needed to combust and produced 6.4 moles of carbon dioxide gas?
1 answer:
You might be interested in
Given :
A compound has a molar mass of 129 g/mol .
Empirical formula of compound is C₂H₅N .
To Find :
The molecular formula of the compound.
Solution :
Empirical mass of compound :
Now, n-factor is :
Multiplying each atom in the formula by 3 , we get :
Molecular Formula, C₆H₁₅N₃
Answer:
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
Best regards!