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

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explanation:

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