Question

When iron(II) chloride reacts with silver nitrate, iron(II) nitrate and silver chloride are produced. The balanced equation for this reaction is: FeCl2 (aq) + 2AgNO3(aq) --> Fe(NO3)2(aq) + 2AgCl(s) Suppose 2.86 moles of iron(II) chloride react. The reaction consumes moles of silver nitrate. The reaction produces moles of iron(II) nitrate and moles of silver chloride. Submit Answer & Next

Answer 1

Answer: The moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles

Explanation:

We are given:

Moles of iron (II) chloride = 2.86 moles

For the given chemical equation:

$FeCl_2(aq.)+2AgNO_3(aq.)\rightarrow Fe(NO_3)_2(aq.)+2AgCl(s)$

• For silver nitrate:

By Stoichiometry of the reaction:

1 mole of iron (II) chloride reacts with 2 moles of silver nitrate.

So, 2.86 moles of iron (II) chloride will react with = $\frac{2}{1}\times 2.86=5.72mol$ of silver nitrate

Moles of silver nitrate reacted = 5.72 moles

• For iron (II) nitrate:

By Stoichiometry of the reaction:

1 mole of iron (II) chloride produces 1 mole of iron (II) nitrate

So, 2.86 moles of iron (II) chloride will produce = $\frac{1}{1}\times 2.86=2.86mol$ of iron (II) nitrate

Moles of iron (II) nitrate produced = 2.86 moles

• For silver chloride:

By Stoichiometry of the reaction:

1 mole of iron (II) chloride produces 2 moles of silver chloride

So, 2.86 moles of iron (II) chloride will produce = $\frac{2}{1}\times 2.86=5.72mol$ of silver chloride

Moles of silver chloride produced = 5.72 moles

Hence, the moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles