Given that the molar mass of h2o is 18.02 g/mol, how many liters of propane are required at stp to produce 75 g of h2o from this reaction?

Given:

H2O, mass 75g of H2O at STP

molar mass of H2O is 18.02 g/mol

Required:

Liters of propane

Solution:

C3H8 + 5O2 -> 3CO2 + 4H2O

get the number of moles of H2O using the molar mass of H2O is 18.02 g/mol

n = 75g/18.02 g/mol = 4.162 moles H2O

change moles H2O to moles propane

4.162 moles H2O(1 mole C3H8/4 moles H2O) = 1.041 moles C3H8

From the ideal gas equation PV = nRT

V = nRT/P

V = (1.041 moles C3H8)(0.08206 L-atm/mol-K) (273K)/(1 atm)

V = 23.3 L of C3H8

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.