Question

Nane CHE I11 3. Calculate the heat released (kl) in the reaction of 0.2021. of acetylene (C2H2) and .05L of hydrogen gas at S


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Answer #1

STP .

P = 1 atm

T = 273 K

moles of C2H2 = P V / R T

                        = 1 x 0.202 / 0.0821 x 273

                         = 9.01 x 10^-3

moles ratio of C2H2 = 9.01 x 10^-3 / 2

                                 = 4.50 x 10^-3

moles of O2 = P V / R T

                   = 1 x 0.05 / 0.0821 x 273

                  = 2.23 x 10^-3

moles ratio of O2 = 2.23 x 10^-3   / 5 = 4.46 x 10^-4

limiting reagent is O2

2C2H2 + 5 O2 -----------------> 4CO2 + 2H2O , DH = -2320 kJ / mol

5 moles O2 -------------------> 2320 kJ

4.46 x 10^-4 mole O2 ----------------------> 2320 x 4.46 x 10^-4 / 5 = 0.207

heat released = 0.207 kJ

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