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Alecsey [184]
2 years ago
15

What is the mole fraction of sulfuric acid in a solution made by adding 34 g of sulfuric acid (H2SO4 = 98 g/mol) to 100 g of wat

er (H2O = 18 g/mol)?
Chemistry
1 answer:
sladkih [1.3K]2 years ago
7 0

Answer:

Mole Fraction  =  0.059

Explanation:

To find the mole fraction, you need to (1) convert grams H₂SO₄ and H₂O to moles (via molar mass) and then (2) calculate the mole fraction.

(Step 1)

Molar Mass (H₂SO₄): 98 g/mol

34 grams H₂SO₄            1 mole
--------------------------  x  -------------------  =  0.347 moles H₂SO₄
                                      98 grams

Molar Mass (H₂O): 18 g/mol

100 grams H₂O              1 mole
-------------------------  x  --------------------  =  5.56 moles H₂O
                                      18 grams

(Step 2)

                                             moles solute
Mole Fraction  =  -------------------------------------------------
                               moles solute + moles solvent

                                              0.347 moles H₂SO₄
Mole Fraction  =  -----------------------------------------------------------
                               0.347 moles H₂SO₄ + 5.56 moles H₂O

Mole Fraction  =  0.059

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Given the partial equation:
Nikolay [14]

Answer : The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

First balance the main element in the reaction.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

Now balance oxygen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-+3H_2O

Now balance hydrogen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-+6H^+\rightarrow I^-+3H_2O

Now balance the charge.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}+2e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : 2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

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