Chemistry, 15.10.2019 04:20 madams1820

A) how many grams of silver chloride can be prepared by the reaction of 107.0 ml of 0.21 m silver nitrate with 107.0 ml of 0.14 m calcium chloride?

g

b) calculate the concentrations of each ion remaining in solution after precipitation is complete.

ag+ m
no3? m
ca2+ m
cl? m

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Answers

Answer from: 885122bah

a.3.2204 g

b.Ag+ = 0 M

NO3? = 0.105 M

Ca2+ = 0.07M

Cl? =0.035 M

Explanation:

The balanced equation for this reaction is:

$2AgNO_{3}+CaCl_{2} - - - 2AgCl + Ca(NO_{3})_{2}$

mol of $AgNO_{3}$

mol= Molarity*volume(L)=0.21M*0.107L=0.02247 mol

mol of $CaCl_{2}$

mol= Molarity*volume(L)=0.14M*0.107L=0.01498 mol

from the reaction stoichiometry we know that we need 2 moles of $AgNO_{3}$ per every mol of $CaCl_{2}$ .

so for 0.01498 mol of $CaCl_{2}$ we need 0.01498*2 =0.02996 mol of $AgNO_{3}$ . Since we have just 0.02247 mol, this will be the reaction limit reagent for this reaction.

we will use all the $AgNO_{3}$

mol of $CaCl_{2}$ used = $\frac{0.02247}{2}=0.011235mol$

For the calculation we always use the limit reagent.

a. How many grams of silver chloride.

From the reaction stoichiometry we know that 2 mol of $AgNO_{3}$ produce 2 mol of AgCl so, 0.02247 mol of $AgNO_{3}$ will the same number of mol of AgCl.

AgCl produced = 0.02247 mol

molecular weight of AgCl = 143.32 g/mol

mass of AgCl = $mol*Mw = 0.02247 mol*143.32 \frac{g}{mol}=3.2204 g$

b.concentrations of each ion

volume of the remained solution 107 mL + 107mL= 214 mL or 0.214 L

$Ag^{+}$ since we used all the $AgNO_{3}$ to produced AgCl and this precipitated we don't have $Ag^{+}$ in the solution after the precipitation.

$NO_{3}$ this ion was used to produce $Ca(NO_{3})_{2}$ and this is soluble in water. The moles of $NO_{3}$ will be 2 times the number of moles of $Ca(NO_{3})_{2}$ produced o simply the same number of moles of $2AgNO_{3}$ before the reaction because this ion did not precipitate but it's concentration change because the volume changed.

concentration = $\frac{0.02247 mol}{0.214L}=0.105 M$

following the same analysis for made for $NO_{3}$ we know that $Ca^{2+}$ did not precipitated so we have the same number of moles of $CaCl_{2}$ in the original sample but in a different volume

concentration = $\frac{0.01498 mol}{0.214L}=0.07 M$

the Cl in the solution after the precipitation will be the Cl that did not reacted since all the Cl that reacted to produce AgCl precipitated.

Cl that reacted is the same number of AgCl produced this is=0.02247 mol

Cl before the reaction is 2 times the number of moles of $CaCl_{2}$ in the sample this is: 0.01498*2 =0.02996 mol

Cl remained = 0.02996 mol -0.02247 mol=0.00749 mol

concentration = $\frac{0.00749 mol}{0.214L}=0.035 M$

Answer from: Quest

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

Answer from: Quest

that tool would be a compass

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