When 3.0 grams of H2 is reacted with excess C at constant pressure, the reaction forms CH4 and releases 53.3 kJ of heat. C(s) + 2 H2(g) --> CH4(g) What is the enthalpy of the reaction in kJ/mol of CH4? Enter your answer numerically and in terms of kJ/mol.

THE ENTHALPY OF REACTION IN KJ/MOL OF CH4 IS 7.07 KJ/MOL.

Explanation:

Mass of H2 = 3 g

Molar mass of H2 = 2 g/mol

Heat released = 53.3 kJ

Equation of the reaction:

C(s) + 2H2(g) > CH4(g)

First:

Calculate the number of moles of H2 that was used:

Number of moles = mass / molar mass

Number of moles = 3g / 2g

Number of moles = 1.5 moles

So therefore, when 53.3 kJ of heat was released from the reaction, 1.5 moles of hydrogen was used.

From the equation of the reaction, one mole of carbon reacts with two moles of hydrogen to form one mole of methane.

For 3 g of hydrogen, 1.5 mole of hydrogen is involved.

It means:

1.5 moles of hydrogen reacts with 0.75 moles of carbon and produces 0.75 moles of methane. This is so because the reaction occurs in 1: 2: 1 in respect to carbon, hydrogen and methane respectively.

So we can say that the production of 0.75 mole of methane will evolve 53.3 kJ of heat.

0.75 mole of methane releases 53.3 kJ of heat.

1 mole of methane will release ( 53.3 kJ * 1 / 0.75 )

= 71.0666 kJ of heat

In conclusion, the enthalpy of the reaction in kJ/ mole of CH4 is 71.07 kJ/mol.

jjj89t'ji; it hink the answer is your mom is not si

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3