Upon balancing the equation below, how many moles of calcium chloride are needed to react completely with 4.3 moles of silver nitrate? AgNO3 + CaCl2 yields AgCl + Ca (NO3) 2

2AgNO₃ (silver-nitrate) + CaCl₂ (calcium-chloride) = 2AgCl (silver-chloride) + Ca (NO₃) ₂ (calcium-nitrate)

n (AgNO₃) = 4,3mol

n (CaCl₂) = ?

From reaction: n (AgNO₃) : n (CaCl₂) = 2 : 1

4,3mol : n (CaCl₂) = 2 : 1, amount of calcium-chloride is two times less than amount of silver-nitrate.

n (CaCl₂) = 2,15 mol