How many grams of iron will be required to release all the antimony from 10 grams of antimony trisulfide? iron 2 sulfide is also formed

mass of Fe = 4.86 g

Explanation:

Balanced chemical equation  

Sb₂S₃ + 3Fe → 2Sb + 3FeS

Given data  from eq.

moles of iron = 3 mol  

mas of iron = ?

moles of antimony trisulfide = 1 mol  

mass of antimony trisulfide = 10 g

Solution  

1st we will find out the mole ratio of Sb₂S₃ and Fe from balanced chemical equation  

            Sb₂S₃ : Fe

                     1 : 3  

Now we find out number of moles of citric acid  Sb₂S₃ from the given mass

           moles = mass / molar mass  

   moles = 10 g / 339.7 g/mol  

   moles = 0.029 mol  

Now we find out moles (x) of iron needed for 0.029 mol  of antimony trisulfide  

from the balanced chemical equation the mole ratios are:  

                   1 : 3

           0.029 : x  

Cross multiply these ratios  

        3 × 0.029 = 1x  

        0.087 = 1 x  

         x = 0.087 mol  

Now we will find out the mass of iron

       mass = moles × molar mass  

       mass of Fe = 0.087 mol × 55.845 g/mol

        mass of Fe = 4.86 g  

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.