Need answer for chem homework asap pls

just need the answer so i can have it show as completed pls

The answer to your question is: kc = 6.48

Explanation:

Data

             Given                 Molecular weight

CaO =    44.6 g                    56 g

CO₂ =     26 g                       44 g

CaCO₃ = 42.3 g                  100 g

Find moles

        CaO                 56 g ----------------  1 mol

                                 44.6 g --------------   x

                               x = (44.6 x 1) / 56 = 0.8 mol

        CO₂                 44 g -----------------  1 mol

                                26 g ----------------    x

                               x = (26 x 1 ) / 44 = 0.6  moles

        CaCO₃            100 g ---------------   1 mol

                                42.3g --------------    x

                              x = (42.3 x 1) / 100 = 0.423 moles

Concentrations

        CaO     =    0.8 / 6.5  = 0.12 M

         CO₂    =     0.6 / 6.5 = 0.09 M

        CaCO₃  =   0.423 / 6.5 = 0.07 M

Equilibrium constant =  

Kc = [0.07] / [[0.12][0.09]

Kc = 0.07 / 0.0108

kc = 6.48

i believe it is c.) 1575 j.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2