The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.06 L of KOH with unknown concentration is placed in a flask with bromothymol blue indicator. A solution of 0.20 M H2SO4 is dripped into the KOH solution. After exactly 0.017 L of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH

The concentration of the KOH is 0.056 M.

Explanation:

The analytical method of titration is performed between KOH and H2SO4 in which bromothymol blue is a pH indicator used to detect the neutral point in a titration.

The formula used in the titration is:

Macid x V acid = Mbase x V base

The volume is in litre.  

Data given:

volume of the base KOH = 0.06 Litre

molarity of the base KOH = ?

volume of the acid = 0.017 L

Molarity of the acid H2SO4 = 0.20M

Applying the formula and putting the values:

0.20 x 0.017 = Mbase x 0.06

M base = 0.20 x 0.017 ÷ 0.06

              = 0.056 M

0.017 litre of  0.20 M acid is added against the 0.06 litres of KOH having molarity of 0.056 M.

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