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3 years ago

How to distinguish a solution of sodium sulphite from a solution of sodium sulphate in the lab

Chemistry

1 answer:

Sonja [21]3 years ago

8 0

We have to distinguish a solution of sodium sulphite from a solution of sodium sulphate in the laboratory.

Solution of sodium sulphite is acidified with dilute HCl and to that when few drops of barium chloride (BaCl₂) solution is added, white precipitate is formed. The white precipitate is soluble in HCl.

To this solution 2 drops of iodine (I₂) solution is added and brown colour of iodine is discharged as I₂ gets reduced to HI.

The reactions involved in case of sodium sulphite is are:

Na₂SO₃ + BaCl₂ = BaSO₃ ↓ + 2NaCl

(white precipitate)

BaSO₃ + 2HCl = BaCl₂ + H₂SO₃

H₂SO₃ + I₂ + H₂O = H₂SO₄ + 2HI

On the other hand, solution of sodium sulphate is acidified with dilute HCl and to that when few drops of barium chloride (BaCl₂) solution is added, white precipitate is formed. The white precipitate of BaSO₄ is formed which is insoluble in HCl.

Na₂SO₄+ BaCl₂ = BaSO₄ ↓ + 2NaCl

(white precipitate)

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In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

$n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}$

$n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{Fe} = 0.5\:mol}$

* to n(S)

$n_{S} = \dfrac{m_{S}}{MM_{S}}$

$n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{S} = 0.5\:mol}$

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

$Fe + S \Longrightarrow FeS$

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

$n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}$

$m_{FeS} = n_{FeS}*MM_{FeS}$

$m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

44 grams of iron sulfide

___________________________

$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

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</span>

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