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polet [3.4K]
1 year ago
13

Liquid methanol (CH₃OH) can be used as an alternative fuel in pickup and SUV engines. An industrial method for preparing it invo

lves the catalytic hydrogenation of carbon monoxide:
CO(g) + 2H₂(g) → CH₃OH(l)
How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 18.5 L of H₂ at 75°C and 744 torr?
Chemistry
1 answer:
r-ruslan [8.4K]1 year ago
7 0

40.6 kJ of heat energy had been emitted.

CO(g) + 2H2(g) CH3OH(l)CO volume, V (CO), equals 15 L or 0.015 m3.

Temperature = 85 0C = 85 + 273 = 358 K Pressure = 112 kPa = 112,000 PaPV = nRT n= 112000 0.015 / 8.314 358 n(CO) = 0.56 moles,

according to the ideal gas law.H2 volume is 14.4 L or 0.0144 m3

T = 750C + 273 K = 348 K n(H2) = 99191.84 0.0144 m3 / 8.314 348 K = 0.49 moles of H2 Pressure = 744 torr = 99191.84 Pa

Hydrogen is the limiting reagent, according to the calculation above.CH3OH = H2 = 0.49/2 = 0.245 m-238.6 (-110.5) = -128.1 kJ/mol for H(rxn) = H(f) (CH3OH) - H (rxn)

We must now multiply H(rxn) by the number of moles of methanol.

E = H(rxn) n(CH3OH) = 128.1 0.245 = 40.6 kJ.

Learn more about Ideal gas law here-

brainly.com/question/13821925

#SPJ4

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A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h
statuscvo [17]

Explanation:

First, calculate the moles of CO_{2} using ideal gas equation as follows.

                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  \frac{mass}{12 g/mol}            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  \frac{0.95 g}{16 g/mol}            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_{3}H_{6}O.

Thus, we can conclude that empirical formula of the given compound is C_{3}H_{6}O.            

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Answer:

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