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NeTakaya
3 years ago
15

Which of the following is an oxidation-reduction reaction? Group of answer choices H2CO3 (aq) + Ca(NO3)2 (aq) → 2HNO3 (aq) + CaC

O3 (s) AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3 (aq) Ba(C2H3O2)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2NaC2H3O2(aq) Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq) HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)
Chemistry
1 answer:
sukhopar [10]3 years ago
8 0

Answer:

Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq) is an oxidation -reduction reaction

Explanation:

An oxidation-reduction reaction is a chemical reaction that involves a transfer of electrons between two atoms, molecles,.. It is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.

⇒The oxidation state of an individual atom is 0.

⇒The total oxidation state of all atoms in: a neutral species is 0 and in an ion is equal to the ion charge.

⇒Group 1 metals have an oxidation state of +1 and Group 2 an oxidation state of +2

⇒Hydrogen generally has an oxidation state of +1 in compounds

⇒Oxygen generally has an oxidation state of -2 in compounds

H2CO3 (aq) + Ca(NO3)2 (aq) → 2HNO3 (aq) + CaCO3 (s)

Carbon on the left has an oxidation state of -4, on the right it has also oxidation state -4

Nitrogen has on the left an oxidation state of +5, on the right it has also an oxidation state of +5

Calcium has on the left an oxidation state of +2, on the right also an oxidation state of +2

⇒ There is no change of oxidation number so this is NOT a redox reaction

AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3

Silver (Ag) on the left has an oxidation state of +1, on the right it has also oxidation state +1

Nitrogen has on the left an oxidation state of +5, on the right it has also an oxidation state of +5

Cl has on the left an oxidation state of -1, on the right it has also an oxidation state of -1

⇒ There is no change of oxidation number so this is NOT a redox reaction

Ba(C2H3O2)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2NaC2H3O2(aq)

Na has on the left an oxidation state of +1, on the right it has also an oxidation state of +1

Ba has on the left an oxidation state of +2, on the right it has also an oxidation state of +2

S has on the left an oxidation state of +6, on the right it has also an oxidation state of +6

⇒ There is no change of oxidation number so this is NOT a redox reaction

Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq)

Cu has on the left an oxidation state of 0, on the right it has an oxidation state of +2

Ag has on the left an oxidation state of +1, on the right it has also an oxidation state of 0

⇒ There is a change of oxidation number so this is a redox reaction

Oxidizing agent: Ag in AgNO3 ; Reducing agent: Cu

HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)

Na has on the left an oxidation state of +1, on the right it has also an oxidation state of +1

Cl has on the left an oxidation state of -1, on the right it has also an oxidation state of -1

Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq) is an oxidation -reduction reaction

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For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
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Molar mass of KClO_{3} = 39.1 + 35.5 + 3(16.0) = 122.6 g

Molar mass of KCl = 39.1 + 35.5 = 74.6 g

Molar mass of O_{2} = 32.0 g

According to the equation, 2 moles of KClO_{3} reacts to give 3 moles of oxygen.

Therefore, 2 (122.6) = 245.2 g of KClO_{3} will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of KClO_{3} gives 96.0 g of oxygen.

(a) Calculate the amount of oxygen given by 2.72 g of KClO_{3} as follows.

       \frac {96g}{245.2g} \times 2.72 g = 1.06 g of O_{2}


(b) Calculate the amount of oxygen given by 0.361 g of KClO_{3} as follows.

    \frac {96g}{245.2g} \times 0.361 g = 0.141 g of O_{2}


c) Calculate the amount of oxygen given by 83.6 kg KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 83.6 g = 32.731 kg of O_{2}

Convert kg into grams as follows.

    \frac{32.731 kg}{1 kg} \times 1000 g = 32731 g of O_{2}


(d) Calculate the amount of oxygen given by 22.5 mg of KClO_{3} as follows.  

    \frac {96g}{245.2g} \times 22.5 mg = 8.79 mg

Convert mg into grams as follows.

  \frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g of O_{2}

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3 years ago
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