Answer:
0.09 g of precipitate can be formed.
Explanation:
The chemical equation can be written as follows:
AgNO₃ + KCl → AgCl(↓) + NO₃⁻ + K⁺
From the equation, we know that 1 mol AgNO₃ reacts with 1 mol KCl to produce 1 mol AgCl.
The problem gives us data to calculate the initial number of moles of silver nitrate and potassium chloride:
n° of moles of silver nitrate = concentration * volume
n° of moles of silver nitrate = 0.10 mol/l * 0.006 l = 6 x 10⁻⁴ mol AgNO₃
n° of moles of KCl = 0.15 mol/l * 0.005 l = 7.5 x 10⁻⁴ mol KCl
Since 1 mol AgNO₃ reacts with 1 mol KCl, 6 x 10⁻⁴ mol AgNO₃ will react with 6 x 10⁻⁴ mol KCl and produce 6 x 10⁻⁴ mol AgCl.
1.5 x 10⁻⁴ mol KCl is in excess.
The molar mass of AgCl is 143.32 g/mol, then, 6 x 10⁻⁴ mol AgCl will have a mass of (6 x 10⁻⁴ mol AgCl * 143.32 g / 1 mol) 0.09 g.
