2so3 (g) --> 2 so2 (g) + o2 (g) you place 0.0300 mol of pure so3 in an 8.00-l flask at 1150 k. at equilibrium, 0.0058 mol of o2 has been formed. calculate k for this reaction at 1150 k. for k, use scientific notation and three significant figures. for scientific notation use "e" to represent "x10^" and no spaces in your numerical answer.

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