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Silver nitrate and iron (III) chloride are reacted. 27.0 g silver nitrate and 43.5 g iron (III) chloride are used in the reaction.

3 AgNO3 + FeCl3 --> 3 AgCl + Fe(NO3)3

1. Determine the limiting reactant for this reaction.

User Timberlake
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Answer:

To determine the limiting reactant, we need to calculate the amount of product that can be formed from each reactant and see which reactant produces the least amount of product. We will use the balanced chemical equation:

3 AgNO3 + FeCl3 --> 3 AgCl + Fe(NO3)3

First, let's calculate the amount of product that can be formed from the silver nitrate:

From the balanced equation, we know that 3 moles of silver nitrate react with 1 mole of iron (III) chloride to produce 3 moles of silver chloride and 1 mole of iron (III) nitrate.

Molar mass of AgNO3 = 107.87 g/mol

Molar mass of AgCl = 143.32 g/mol

Using the given mass of silver nitrate:

27.0 g AgNO3 x (1 mol AgNO3 / 107.87 g AgNO3) x (3 mol AgCl / 3 mol AgNO3) x (143.32 g AgCl / 1 mol AgCl) = 92.9 g AgCl

Next, let's calculate the amount of product that can be formed from the iron (III) chloride:

Molar mass of FeCl3 = 162.20 g/mol

Molar mass of Fe(NO3)3 = 241.81 g/mol

Using the given mass of iron (III) chloride:

43.5 g FeCl3 x (1 mol FeCl3 / 162.20 g FeCl3) x (3 mol AgCl / 1 mol FeCl3) x (143.32 g AgCl / 1 mol AgCl) = 104.6 g AgCl

Comparing the two amounts of product, we can see that the amount of AgCl that can be formed from the iron (III) chloride is greater than the amount that can be formed from the silver nitrate. Therefore, silver nitrate is the limiting reactant in this reaction.

Step-by-step explanation:

User Nashir Uddin
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