I have post this question 3 times the answers were wrong 1010kJ, -168kJ and -197

Question

I have post this question 3 times the answers were wrong 1010kJ, -168kJ and -19734.88kJ

± Enthalpy of Reaction: State and Stoichiometry

Use the data below to answer the questions.

Keep in mind that the enthalpy of formation of an element in its standard state is zero.

Part D

Suppose that 0.290 mol of methane, CH4(g), is reacted with 0.440 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole products. Assuming that the reaction occurs at constant pressure, how much heat is released?

Express your answer to three significant figures and include the appropriate units

Substance Hf (kJ/mol) C(g) 718.4 CF4(g) 679.9 CH4(g) 74.8 H(g) 217.94 HF(g) 268.61

Explanation / Answer

CH4 + F2 --> CF4 + HF

Balance equations

CH4 + 4F2 --> CF4 + 4HF

Calculate enthalpy

H = Hprod - Hreact = (679.9 + 4*-268.61) - (-74.8 + 4*0) = -1679.54 kJ/rxn

If 0.29 mol of CH4 react with 0.44 mol o F2, find limitng reactant

ratio is 1:4

therefore F2 is limiting

0.44 mol of F2 will react with 0.44/4 = 0.11 mol o0f CH4

the basis:

-1679.54 kJ/rxn --> 1 mol of CH4

but we have 0.11 mol so

0.11*(-1679.54) = -184.7494 kJ

Q = 184,75 kJ of heat are released

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