Chemistry help ASAP? Chemical Kinetics? please explain with work if possible i just don't understand.?

the general rate equation is
.. rate = k * [A]^x * [B]^y * [C]^z

where x, y, z are the "orders" with respect to A, B, and C

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you are given
.. x = 2.. (that's what "2nd order in A" means)
.. y = 1.. (that's what "1st order in B" means)
and
.. z = 0.. (think about it.. C^0 = 1 so that the rate equation can be re-written as
.. .. .. .. . .rate = k * [A]² * [B]¹ * [C]⁰ = k * [A]² * [B]¹ * 1 = k * [A]² * [B]¹
.. .. .. ... ..meaning the rate is independent of the concentration of C
... .. .. . .. that was given in the problem statement... REMEMBER this
.. .. . .. .. .if the rate is independent of concentration of component X
.. .. . .. .. .the order with respect to X = 0

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now we know
.. rate = k * [A]² * [B]

and then
.. k = rate / ([A]² * [B])

plugging in the data
.. k = (0.08 mol / L•sec) / ( (3 mol / 3L)² * (6 mol / 3L) ) = 0.04 L² / mol²•sec)

or if you prefer
.. k = 0.04 L² mol⁻² sec⁻¹

2A(g) + 3B(g) + 5C(g) → Products

The reaction is found to be 2nd order in the concentration of A, 1st order in the concentration of B, and independent of the concentration of C. Then,
(Initial Rate) = k [A]² [B]

When [A] = (3 mol)/(3 L) = 1 mol*L⁻¹ and [B] = (6 mol)/(3 L) = 2 mol*L⁻¹, (Initial rate) = 0.08 mol*L⁻¹*s⁻¹ :
0.08 mol*L⁻¹*s⁻¹ = k (1 mol*L⁻¹) (2 mol*L⁻¹)
Then, k = (0.08 mol*L⁻¹*s⁻¹) / {(1 mol*L⁻¹) (2 mol*L⁻¹)} = 0.04 mol⁻¹*L*s⁻¹

The reaction rate constant, k = 0.04 mol⁻¹*L*s⁻¹