Determine the mass (in grams) of silver chloride, AgCl, that would be precipitated if an excess of AgNO3 solution is added to 55.0 cm3 of 0.200 mol dm−3 KCl solution. The balanced equation for the reaction is:KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)Mr AgCl = 143.32
Question
Determine the mass (in grams) of silver chloride, AgCl, that would be precipitated if an excess of AgNO3 solution is added to 55.0 cm3 of 0.200 mol dm−3 KCl solution. The balanced equation for the reaction is:KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)Mr AgCl = 143.32
Solution
Step 1: Determine the number of moles of KCl in the solution.
The number of moles is given by the formula: Moles = Concentration (mol/dm^3) * Volume (dm^3)
Given that the volume is in cm^3, we need to convert it to dm^3 by dividing by 1000.
So, Moles of KCl = 0.200 mol/dm^3 * 55.0 cm^3 / 1000 = 0.011 mol
Step 2: Determine the number of moles of AgCl that will be precipitated.
From the balanced chemical equation, we can see that one mole of KCl reacts with one mole of AgNO3 to produce one mole of AgCl. Therefore, the number of moles of AgCl that will be precipitated is equal to the number of moles of KCl in the solution.
So, Moles of AgCl = 0.011 mol
Step 3: Determine the mass of AgCl that will be precipitated.
The mass is given by the formula: Mass = Moles * Molar mass
So, Mass of AgCl = 0.011 mol * 143.32 g/mol = 1.57652 g
Therefore, the mass of AgCl that would be precipitated is approximately 1.58 g.
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