Kim found that 10.0 mL of a 0.50 M NaOH solution are required to completely titrate 25,0 mL of HCl solution. What is

the concentration of the HCl?

0.20M HCl

Explanation:

The reaction of neutralization of NaOH with HCl is:

NaOH + HCl → H₂O + Cl⁻ + Na⁺

Where 1 mole of NaOH reacts with 1 mole of HCl.

Moles of 10.0 mL of a 0.50 M NaOH solution are:

0.0100L × (0.50moles NaOH / 1L) = 0.0050 moles of NaOH

As 1 mole of NaOH reacts with 1 mole of HCl, at equivalence point, moles of HCl are 0.0050 moles.

If volume of HCl is 25.0mL, concentration is:

0.0050 mol / 0.025L = 0.20M HCl

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