Consider the following reaction: CH4 +2O2 → CO2 + 2H2O. ΔH= -2891 kJ

Calculate the enthalpy change for each of the following cases:

a. 1.00 g methane is burned in excess oxygen.
b. 1.00 3 10^3 L methane gas at 740. torr and 258°C are burned in excess oxygen.

(a)

(b)

Explanation:

Hello.

(a) in this case, since the change in the molar enthalpy for this reaction is ΔH= -2891 kJ per mole of methane, we can compute the total change in the enthalpy by considering the given mass of methane which should be converted to moles as shown below:

(b) For this one, we first compute the moles of methane by using the ideal gas equation:

Then, we compute the change in the enthalpy as done on (a):

Regards.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

a electronegativity decreases beacuse the size of nucleus is bigger.