How many moles of precipitate will be formed when 50.0 mL of 0.300 M AgNO3 is reacted with excess CaI2 in the following chemical reaction? 2AgNO3 (aq) + CaI2 (aq) - -> AgI (s) + Ca (NO3) 2 (aq)

0.015 mol AgI

Explanation:

First of all, you have provided an unbalanced equation, so let's balance that:

2AgNO3 (aq) + CaI2 (aq) - -> 2AgI (s) + Ca (NO3) 2 (aq)

Then let's calculate the moles of AgNO3 using the formula: n = c * V:

c (AgNO3) = 0.300 M (mol/L)

V (AgNO3 - solution) = 50.0 mL = 0.05 L

n (AgNO3) = c (AgNO3) * V (AgNO3) = 0.300 mol/L * 0.05 L = 0.015 mol

Finally, using the coefficients in the equation, we find the moles of AgI:

n (AgNO3) : n (AgI) = 1 : 1

n (AgI) = n (AgNO3) = 0.015 mol