Question

Consider the combustion of hydrochloric acid. 4 HCl (aq) + O2(g)  2H2O(l) + 2 Cl2(g) ΔHrxn = + 96.976 kJ/mol

B.If you place 750 mL of 0.500 M solution of HCl with 17.2 g of O2, will any oxygen remain? (Hint: limiting reagent problem)

C.What mass of chlorine gas (in grams) is produced?

D) How many molecules of Cl2 are produced based on part (B)

E) What mass of excess reagent remains after the limiting reagent is used up?

F) If the chemist collects 6.98 g of Cl2, what is the percent yield of the reaction? What is the percent error?

G) What is the enthalpy of the reaction for 24.5 g of HCl in presence of excess O2? Was energy absorbed or released into the surroundings?

Answer 1

the reaction is

4 HCl + 02 ---> 2H20 + 2 Cl2

we know that

moles = molarity x volume (L)

so

moles of HCl = 0.5 x 0.75 = 0.375

also

moles = mass / molar mass

so

moles of 02 = 17.2 / 32 = 0.5375

now

4 HCl + 02 ---> 2H20 + 2 Cl2

moles of 02 required = ( 1/4) x moles of HCl

moles of 02 required = 0.25 x 0.375

moles of 02 required = 0.09375

but

0.5375 moles of 02 are present

so

02 is in excess , HCl is the limiting reagent

now

E)

moles of 02 left = 0.5375 - 0.09375 = 0.44375

mass of 02 left = 0.44375 x 32 = 14.2

so

14.2 grams of 02 is in excess

C)

now

4 HCl + 02 ---> 2H20 + 2 Cl2

moles of Cl2 formed = 0.5 x moles of HCl reacted

moles of Cl2 formed = 0.5 x 0.375

moles of Cl2 formed = 0.1875

mass of Cl2 fomred = 0.1875 x 71

mass of Cl2 produced = 13.3125 grams

D)

we know that

number of molecules = moles x avagadro number

so

number of Cl2 molecules = 0.1875 x 6.023 x 10^23

number of Cl2 molecules = 1.129 x 10^23

F)

% yield = actual x 100 / theoretical

% yield = 6.98 x 100 / 13.3125

% yield = 52.43

G)

now

moles = mass / molar mass

so

moles of HCl = 24.5 / 36.5

moles of HCl = 0.6712

now

4 moles of HCl --->   96.976 kJ

0.6712 moles of HCl ---> y kJ

y = 0.6712 x 96.976 / 4

y = 16.273

so

16.273 kJ is the enthalpy

it has positive sign

so

energy is absorbed from the surroundings