For a given sample of ch3oh, the enthalpy change during the reaction is 82.3 kj . how many grams of methane gas are produced?

The equation is as follows;
 2CH3OH(g) = 2CH4(g) + O2(g), ΔH= +252.8 kJ
From the equation; for the reaction to produce 2 moles of methane (32g) an energy of 252.8 kJ is released.
Therefore; for an energy of 82.3 kJ the number of moles that will be produced will be;    =  (2×82.3)/252.3
                               = 0.6524 moles
                  which is equivalent to 0.6524 × 16 = 10.438 g
There, the mass of CH4 produced will be 10.438 g

The complete question in the attached figurewe have that3^y = 8we can apply base 3 logarithm to both sides.  base 3 log (3^y) = base 3 log (8)y * [base 3 log 3] = base 3 log (8)we know thatbase 3 log 3 = 1therefore  y = base 3 log (8)the answer is  y = base 3 log (8)