All the iron in a 2.000 g tablet was dissolved in an acidic solution and converted to Fe2+. This was then titrated with KMnO4. The titration required 27.50 cm3 of 0.100 mol dm-3 KMnO4. Calculate the total mass of iron in the tablet and its percentage by mass.

0.768g of Fe are in the tablet = 38.4% by mass of Fe in the tablet

Explanation:

The balanced redox equation of Fe²⁺ with KMnO₄ in acidic medium is:

5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Where 5 moles of Fe²⁺ react per mole of KMnO₄

The moles required of KMnO₄ are:

27.50cm³ * (1dm³ / 1000cm³) = 0.0275dm³ * (0.100mol / dm³) = 0.00275mol KMnO₄

Moles of Fe²⁺ are:

0.00275mol KMnO₄ * (5 moles Fe²⁺ / 1mol KMnO₄) = 0.01375moles of Fe²⁺

In grams (Molar mass Fe: 55.845g/mol):

0.01375moles of Fe²⁺ * (55.845g/mol) =

And percentage is:

0.768g Fe / 2.000g * 100 =

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.