# Chemistry: Limiting Reagent and Enthalpy Sample Questions

I am having trouble setting up the follow problems. I am not even sure where to start. If anyone can offer any advice, I would greatly appreciate it, as I am completely stumped:

1) The volume in Liters of H2(g), measured at 22 degrees C and 745 mmHg, required to react with 30.0 L of CO (g), measured at 0 degrees C and 760 mmHg, in the reaction 3CO(g)+7H2(g) --> C3H8(g)+3H2O(l) is?

2) The complete combustion of propane, C3H8(g), is represented by the equation C3H8(g)+5O2(g)-->3CO2(g)+4H2O(l). Delta H (change of enthalpy)=-2220 kJ.

How much heat is involved in the complete combustion of 12.5L C3H8(g) at 25 degrees C and 790 mmHg?

https://brainmass.com/chemistry/energetics-and-thermodynamics/chemistry-limiting-reagent-enthalpy-sample-questions-438585

## SOLUTION This solution is **FREE** courtesy of BrainMass!

The volume in Liters of H2(g), measured at 22 degrees C and 745 mmHg, required to react with 30.0 L of CO (g), measured at 0 degrees C and 760 mmHg, in the reaction 3CO(g)+7H2(g) --> C3H8(g)+3H2O(l) is?

First step is to determine the number of moles of H2(g) then you can determine the volume using the Ideal Gas Law.

3 CO2 (g) + 7 H2 (g) →yields C3H8 (g) + 3 H2O (l)

moles CO2 = (P*V)/(R*T) = ( (760 mm Hg)/(760 mm Hg)⁄(1 atm)) *30.0 L)/(0.0820575 (L*atm)⁄(mol*K)*(0+273.15)K) =1.34 moles CO_(2 )

moles H2 (g)=moles CO2 (g)* (4 moles of H_2 (g))/(3 moles of CO2 (g)) = 1.34 moles CO2 (g)* (4 moles of H2 (g))/(3 moles of CO2 (g)) =1.00 moles H2 (g)

V H_2 (g)= (n*R*T)/P = (1.00 moles H_2 (g)* 0.0820575 (L*atm)⁄(mol*K)*(22+273.15)K)/((745 mm Hg)/(760 (mm Hg)⁄(1 atm))) = 24.8 L H2 (g)

The complete combustion of propane, C3H8(g), is represented by the equation C3H8(g)+5 O2(g)-->3 CO2(g) + 4 H2O(l). Delta H (change of enthalpy)=-2220 kJ.

How much heat is involved in the complete combustion of 12.5L C3H8(g) at 25 degrees C and 790 mmHg?

From the Heat of Enthalpy we get the equation

3 C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g)

moles C_3 H_8 (g) = (P*V)/(R*T) =((790 mm Hg)/(760 (mm Hg)⁄(1 atm)) *12.5 L)/(0.0820575 (L*atm)⁄(mol*K)*(25+273.15) K = 0.531 moles C_3 H_8 (g)

∆E rxn = -2200 KJ⁄mol* # mol = -2200 KJ⁄mol * 0.531 mol =-1168 KJ

https://brainmass.com/chemistry/energetics-and-thermodynamics/chemistry-limiting-reagent-enthalpy-sample-questions-438585