Chemistry, 27.08.2019 17:10 random456

Suppose 0.701g of iron(ii) chloride is dissolved in 50.ml of a 55.0mm aqueous solution of silver nitrate.
calculate the final molarity of chloride anion in the solution. you can assume the volume of the solution doesn't change when the iron(ii) chloride is dissolved in it.
be sure your answer has the correct number of significant digits.

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Answer from: fish64

Answer : The molarity of chloride anion in the solution is 0.05 mole/L

Explanation : Given,

Mass of FeCl_2 = 0.701 g

Volume of solution = 50 ml = 0.050 L

Molarity of AgCl = 55.0 mM = 0.055 M

Molar mass of FeCl_2 = 126.751 g/mole

First we have to calculate the moles of FeCl_2 .

$\text{Moles of }FeCl_2=\frac{\text{Mass of }FeCl_2}{\text{Molar mass of }FeCl_2}=\frac{0.701g}{126.751g/mole}=0.00553moles$

Now we have to calculate the moles of AgNO_3 .

$\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}=0.055mole/L\times 0.050L=0.00275mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$FeCl_2(aq)+2AgNO_3(aq)\rightarrow 2AgCl(s)+Fe(NO_3)_2(aq)$

From the balanced reaction we conclude that

As, 2 moles of AgNO_3 react with 1 mole of FeCl_2

So, 0.00275 moles of AgNO_3 react with $\frac{0.00275}{2}=0.001375$ moles of FeCl_2

From this we conclude that, FeCl_2 is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgCl .

As, 2 moles of AgNO_3 react to give 2 moles of AgCl

So, 0.00275 moles of AgNO_3 react to give 0.00275 moles of AgCl

Now we have to calculate the molarity of AgCl .

$\text{Molarity of }AgCl=\frac{\text{Moles of }AgCl}{\text{Volume of solution}}$

$\text{Molarity of }AgCl=\frac{0.00275mole}{0.055L}=0.05mole/L$

As we know that, 1 mole of AgCl in solution gives 1 mole of silver ion and 1 mole of chloride ion.

So, the molarity of chloride ion = Molarity of AgCl = 0.05 mole/L

Therefore, the molarity of chloride anion in the solution is 0.05 mole/L

Answer from: Quest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

Answer from: Quest

Ci think not