Answer :
Upon complete reaction of 28.7 grams of silver nitrate with excess iron(II) chloride, approximately 10.12 grams of iron(II) nitrate would be formed, option (e) is correct.
The question is asking how many grams of iron(II) nitrate will be formed when 28.7 grams of silver nitrate react completely with excess iron(II) chloride. To solve this, we use stoichiometry to convert the mass of the reactant (silver nitrate) into moles and then relate it to the molar ratio of the products in the balanced equation for the reaction.
The chemical equation for the reaction between iron(II) chloride and silver nitrate is:
3 AgNO3 + FeCl2 → 3 AgCl + Fe(NO3)2
Using molar masses, we find:
Moles of AgNO3 = mass / molar mass = 28.7 g / 169.87 g/mol
Moles of Fe(NO3)2 produced = Moles of AgNO3 × (1 mol Fe(NO3)2 / 3 mol AgNO3)
Grams of Fe(NO3)2 = Moles of Fe(NO3)2 × molar mass of Fe(NO3)2
Performing the calculations:
Moles of AgNO3 = 28.7 g / 169.87 g/mol ≈ 0.169 mol
Moles of Fe(NO3)2 = 0.169 mol × (1/3) = 0.0563 mol
Grams of Fe(NO3)2 = 0.0563 mol × 179.85 g/mol ≈ 10.12 g
Therefore, the mass of iron(II) nitrate formed will be approximately 10.12 grams, which is option (e).
The Question is incomplete. The complete question is
How many grams of iron(II) nitrate will be formed upon the complete reaction of 28.7 grams of silver nitrate with excess iron(II) chloride?
a) 36.9 grams
b) 73.8 grams
c) 49.2 grams
d) 14.7 grams
e) 10.2 grams