For the following reaction, 45.6 grams of iron(II) chloride are allowed to react with 110 grams of silver nitrate .

iron(II) chloride(aq) + silver nitrate(aq) > iron(II) nitrate(aq) + silver chloride(s)

What is the maximum amount of iron(II) nitrate that can be formed? grams

See explanation below

Explanation:

To know this, let's write again the reaction but with the chemical symbols:

FeCl₂ + 2AgNO₃ > Fe(NO₃)₂ + 2AgCl

Now that we have the reaction and it's balanced, to know the amount of iron nitrate produced, we need to calculate the limiting reactant, and for that, we need the molar mass to calculate the moles.

The reported molar masses are:

MM FeCl₂ = 126.75 g/mol

MM AgNO₃ = 169.87 g/mol

MM Fe(NO₃)₂ = 179.86 g/mol

Now the moles:

moles FeCl₂ = 45.6/126.75 = 0.36 moles

moles AgNO₃ = 110/169.87 = 0.65 moles

Let's calculate now the limiting reactant:

1 mole of FeCl₂ > 2 moles AgNO₃

0.36 moles > X

X = 0.36*2 / 1 = 0.72 moles of AgNO₃

However, we only have 0.65 moles of AgNO₃, therefore, this is the limiting reactant.

As AgNO₃ is the limiting reactant, this will be the moles that will be consumed in the reaction, so the moles of Iron nitrate will be:

moles Fe(NO₃)₂ = 0.65 / 2 = 0.325 moles

Finally the mass:

m = 0.325 * 179.86

m = 58.45 g

iron and sleel

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