Silver nitrate reacts with iron (III) chloride to produce silver chloride and iron (III) nitrate. In a particular experiment, a solution containing 25.0 g of silver nitrate is completely reacted. Write the balanced chemical equation for the reaction and determine how many grams of iron (III) nitrate were formed.

Explanation:

Silver nitrate + iron (III) chloride  --> silver chloride + iron (III) nitrate

The balanced chemical equation is given as;

2FeCl3 + 3Pb(NO3)2 → 2Fe(NO3)3 + 3PbCl2  

Mass of Pb(NO3)2  = 25g

How many grams of iron (III) nitrate were formed?

From the equation;

3 mol of Pb(NO3)2 produces 2 mol of Fe(NO3)3

Converting to mass using;

Mass = Number of moles * Molar mass

Pb(NO3)2;

Mass = 3 mol * 331.2 g/mol

Mass = 993.6 g

Fe(NO3)3;

Mass = 2 mol * 241.86 g/mol

Mass = 483.72 g

This means 993.6 g produces 483.72 g

We have;

993.6 g  = 483.72 g

25 g = x

solving for x;

x = 25 * 483.72  / 993.6

x = 12.1709 g of iron (III) nitrate

sry but any answer choices

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