Question

Suppose that 0.490 mol of methane, CH4(g), is reacted with 0.640 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

Part D Suppose that 0.490 mol of methane, CH4 (g), is reacted with 0.640 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. View Available Hint(s) Value Units heat Submit

Answer 1

The balanced equation is

CH4(g) + 4F2(g) ------------> + 4 HF (g) +CF4(g)

For this rxn Delta Hf = [sum of Hf of products- sum of Heatf of reactants]

= [ 1x(-679.9)+4(-268.61) ] - [ 1(-74.8) + 4(0)]

= -1679.54 kJ

This is the heat given out when 1 moles of CH4(16g/mol) reacted with 4 moles of F2(4x19g/mol)  

however we have taken different moles

CH4(g) + 4F2(g) ------------>   4 HF (g) + CF4(g)

0.490 0.640 0 0 initial moles

To calculate the limiting reagent , let us find the ratio of given to required moles.

0.490/1= 0.49 0.640/4= 0.16

Thus F2 being the smaller ratio, it is the lmiting reagent.

So heat produced is calculated using F2 moles only .

If 4 moles of F2 is reacted we get -1679.54 kJ heat

If 0.640 moles of F2 reacts cmpletely the heat given out = -1679.54kJ x0.640mol /4 mol =-268.73 kJ