Answer:
The enthalpy change for the given reaction is 424 kJ.
Explanation:
We have :
Enthalpy changes of formation of following s:
(standard state)
![\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5B%5CDelta%20H_f%28product%29%5D-%5Csum%20%5B%5CDelta%20H_f%28reactant%29%5D)
The equation for the enthalpy change of the given reaction is:
=
=
The enthalpy change for the given reaction is 424 kJ.
Answer:
First confirm the reaction is balanced:
C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).
a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.
b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).
There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.
MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g
c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.
The other questions use the same technique and will give you some much needed practice.
Explanation:
PV = nRT
P = (nRT)/V
P = (0.3 mol × 0.08206 atm-l/(mol-K) × (273.15 + 30) K)/(0.5 l)
P = 14.9258934 atm
Since a water molecule is H2O, you would divide 126 hydrogen molecules by 2, and you would get 63. That means you have 63 double hydrogen molecules, and 58 oxygen molecules to pair up with them. So that means you could have 58 molecules of water, with 5 double hydrogen molecules, so basically 10 extra molecules of hydrogen along with the H2O molecules. Hope I helped! :)
Answer:
Explanation:
Hello!
In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
Best regards!
