Which substance is the oxidizing agent in the reaction 2CuO+C→2Cu+CO2 ? Express your answer as a chemical formula.

2 answers:

7 0

The chemical reaction is:

<span>2CuO+C→2Cu+CO2

The oxidizing agent is the one that is being reduced which is copper from a charge of +2 to a neutral charge or zero. The opposite is called the reducing agent which is carbon for this case.
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Jobisdone [24]2 years ago

4 0

<u>Answer:</u> The oxidizing agent in the given reaction is copper.

<u>Explanation:</u>

Oxidation reaction is the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

$X\rightarrow X^{n+}+ne^-$

Reducing agents are the agents that reduces the other substance and itself gets oxidized. They undergoes reduction reactions.

Reduction reaction is the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

$X^{n+}+ne^-\rightarrow X$

Oxidizing agents are the agents that oxidizes the other substance and itself gets reduced. These agents undergoes reduction reactions.

For the given chemical reaction:

$2CuO+C\rightarrow 2Cu+CO_2$

The half reactions for the above reaction are:

Oxidation half reaction: $C\rightarrow C^{4+}+4e^-$

Reduction half reaction: $2Cu^{2+}(aq.)+4e^-\rightarrow 2Cu(s)$

From the above reactions, copper is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

Carbon is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Hence, the oxidizing agent in the given reaction is copper.

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Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal

fiasKO [112]

Answer:

1. 3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

$\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}$

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

$\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

(c) Calculate the mass of Na₂CO₃·10H₂O

$\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?; c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}$