What is the ka for 0.03M Solution of HClO4 for 1.90 PH value. A) 7.2x10^-3
B) 8.27x10^-3
C) 8.27x10^-4
D) 6.27x10^-4

(I keep getting 5x10^-3 when I round all the values to 2 decimal places but it’s not in the options)

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