A 25.0 mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. The H3O+ concentration after the addition of 66.2 mL of KOH is M. a. 0.723
b. 1.00 × 10-7
c. 2.81 × 10-13
d. 0.439
e. 0.273

lol idk man i guess u gotta ask bill nye the science guy

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