How many grams of precipitate can be formed in the reaction between 6.00 mL of 0.10 M silver nitrate and 5.00 mL of 0.15 M potassium chloride? Hint: Start with a balanced chemical equation. Do not enter units with your 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.