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

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2