A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions after the equivalence point. Write down the items to their respective table.
i) 5.00 mL of 1.00 M NaOH
ii) 50.0 mL of 1.00 M NaOH
iii) 100 mL of 1.00 M NaOH
iv) 150 mL of 1.00 M NaOH
v) 200 mL of 1.00 M NaOH
O Before equivalence point O At equivalence point O After equivalence point

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

12 is it do you think so