Question

If 37.5 mL of 0.100 M calcium chloride reacts completely with aqueous silver nitrate, what is the mass of AgCl (MM - 143.32 g/mol) precipitate?

CaCl₂(aq) +_ AgNO₃(aq) → __AgCl(s) + __Ca(NO₃)₂(aq)
a) 3.81 g
b) 4.78 g
c) 5.67 g
d) 6.42 g

Answer 1

To determine the mass of AgCl precipitate formed when 37.5 mL of 0.100 M calcium chloride reacts completely with aqueous silver nitrate, calculate the number of moles of AgCl precipitate formed. Then, use the molar mass of AgCl to determine its mass.

Step-by-step explanation:

To determine the mass of AgCl precipitate formed when 37.5 mL of 0.100 M calcium chloride reacts completely with aqueous silver nitrate, we need to calculate the number of moles of AgCl precipitate formed. Based on the balanced chemical equation, 1 mole of CaCl₂ reacts with 2 moles of AgCl. Since the volume of the calcium chloride solution is given and the concentration is known, we can calculate the number of moles of CaCl₂ used. Then, using the mole ratio, we can determine the moles of AgCl precipitate formed and finally calculate its mass using the molar mass of AgCl.

Calculations:

Number of moles of CaCl₂ = volume of solution (L) × concentration (M)

= 0.0375 L × 0.100 mol/L

= 0.00375 mol

Number of moles of AgCl = 2 × number of moles of CaCl₂

= 2 × 0.00375 mol

= 0.0075 mol

Mass of AgCl = number of moles × molar mass

= 0.0075 mol × 143.32 g/mol

= 1.074 g

Therefore, the mass of AgCl precipitate formed is approximately 1.074 g.