Which of the follow which of the following mixtures would result in a buffered solution? a. mixing 100.0 ml of 0.100 m hcl with 100.0 ml of 0.100 m naoh. b. mixing 100.0 ml of 0.100 m nh (k) = 1.8 x 10-5) with 100.0 ml of 0.100 m naoh. c. mixing 100.0 ml of 0.10 m hcl with 100.0 ml of 0.100 m nh3(k) = 1.8 x 10-5). d. mixing 50.0 ml of 0.100 m hcl with 100.0 ml of 0.100 m nh (kb = 1.8 x 10-5).

its not a choice. you dont choose if you want to be gay or straight.

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

i think it's letter a.