The combustion of 0.25 mol of an unknown organic compound results in the release of 320 kj of energy. which of the compounds in the table could be the unknown compound?
a. methane
b. ethylene
c. ethanol

C. ethanol

The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.

The enthalpy of combustion of the unknown compound is

ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol

To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,

e% =  ( | ΔHx - ΔH | / ΔHx ) x 100%

where ΔHx is the enthalpy of combustion of the probable compound.

The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol

CompoundEnthalpy of combustion (kJ/mol)   Deviation

Methane                        - 890.7                                 43.8%

Ehylene                         -1411.2                                   9.3%

Ethanol                        -1368.6                                    6.5%

According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2