Chemistry, 14.04.2020 08:49, Jadaflournoy5

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)

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answered: awuekure

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

answered: Guest

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answered: Guest

1.consider the following balanced equation and answer parts a through. show calculations for each part.

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.

explanation:

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How many moles of precipitate will be formed when 50.0 mL of 0.300 M AgNO3 is reacted with excess Ca...