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Mars2501 [29]
3 years ago
7

Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients

of all species in the balanced equation. (Remember the coefficients that are equal to one.) The sum of the coefficients is: BrO3– (aq) + Sb3+(aq) → Br–(aq) + Sb5+(aq) (acidic solution)
Chemistry
1 answer:
Elena L [17]3 years ago
7 0

Answer : The balanced chemical equation in a acidic solution is,

BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)

The sum of the coefficients is, 17

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,

BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • First balance the main element in the reaction.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • Now balance oxygen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-+3H_2O

  • Now balance hydrogen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-+6H^+\rightarrow Br^-+3H_2O

  • Now balance the charge.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}+2e^-

Reduction : BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : 3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-

Reduction : BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O

The balanced chemical equation in acidic medium will be,

BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

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Which is the correctly balanced chemical equation for the reaction of KOH and H2SO4?
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Suppose you begin with 1.50~g of the hydrate copper(II)sulfate · x-hydrate (CuSO4· x H2O), where x is an integer. After dehydrat
chubhunter [2.5K]

These are two questions and two answers.

Question 1.

Answer: x = 5

Explanation:

1) Data:

m₁ = 1.50 g

compound₁: CuSO₄· x H₂O

m₂ = 0.96g

compound₂ = CuSO₄

x = ? (round to the nearest integer)

2) Solution:

i) molar mass of CuSO₄ = 63.546g/mol + 32.065g/mol + 4×15.999g/mol = 159.607g/mol

ii) number of moles of CuSO₄

number of moles = mass in grams / molar mass = 1.50 g/ 159.607 g/mol = 0.006265 mol

iii) molar mass of H₂O = 18.015 g/mol

iv) mass of H₂O = 1.50g - 0.96g = 0.54g

v) number of moles of H₂O = mass in grams / molar mass = 0.54 g / 18.015 g/mol = 0.0300 mol

vi) Ratio moles H₂O / moles CuSO₄ = 0.0300 / 0.0062625 ≈ 5

∴ x = 5.

Question 2.

Answer: 5.5 g

1) Data:

compound₁ = KAl(SO₄)₂ · 12H₂O.

compound₂ = KAl(SO₄)₂

m₂ = 3.0 g KAl(SO₄)₂

m₁ = ? (two significant figures)

2) Solution:

i) molar mass of KAl(SO₄)₂ = 39.098g/mol + 26.982g/mol + 2×32.065g/mol + 8×15.999g/mol = 258.202

ii) number of moles of KAl(SO₄)₂ = mass in grams / molar mass = 3.0g / 258.202 = 0.011619 mol

iii) number of moles of H₂O = 12 × number of moles of KAl(SO₄)₂ = 12 × 0.011619*12 mol = 0.1394 moles

iv) mass of H₂O = number of moles × molar mass = 0.1394 moles × 18.015 g/mol = 2.5 g (rounded to two significant figures)

v) mass of the original compound = mass of KAl(SO₄)₂ + mass of H₂O = 5.5g

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