A55.0g sample of iron (iii) filings is reacted with 23.8g of powdered sulfur (s8). how much iron (iii) sulfide in moles would be produced in this reaction?

equation:

convert to moles
of iron:

convert to moles
of sulfur:

calculate the
limiting reagent:

solve the problem:

0.74 moles iron (III) sulfide

Explanation:

From the balanced equation of reaction:

1 mole of Fe reacts with 1 mole of S to give 1 mole of FeS.

moles =

mole of Fe = 55/55.8 = 0.99 moles

mole of S = 23.8/32.07 = 0.74 moles

Sulfur is limited in quantity and will therefore determine the rate of reaction.

1 mole of sulfur gives 1 mole of FeS

0.74 moles of sulfur will therefore give 0.74 moles of FeS.

0.74 moles iron (III) sulfide will be produced.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

i believe it is-

d.) gravity pulled particles of dust and gas together to form planets.