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2 years ago

2Na2O2 + 2CO2 → 2Na2CO3 + O2

Chemistry

2 answers:

ahrayia [7]2 years ago

7 0

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂

molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield x 100%

81.0 % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

beks73 [17]2 years ago

3 0

Answer:

8.58 g.

Explanation:

To get the actual yield of a reaction, we can use the relation:

The percent yield = [actual yield]/[theoretical yield] x 100.

The percent yield = 81.0 %.

We need to calculate the theoretical yield:

From the balanced reaction: 2Na₂O₂ + 2CO₂ → 2Na₂CO₃ + O₂.

It is clear that 2.0 moles of Na₂O₂ react with 2.0 moles of CO₂ to produce 2.0 moles of Na₂CO₃ and 1.0 mole of O₂.

The no. of moles of reacted 7.80 g of Na₂O₂ = mass/molar mass = (7.80 g)/(77.98 g/mol) = 0.1 mol.

We can get the no. of moles of produced Na₂CO₃ using cross multiplication:

∵ 2.0 moles of Na₂O₂ produce → 2.0 moles Na₂CO₃.

∴ 0.1 mol of Na₂O₂ produce → 0.1 mol Na₂CO₃.

We can get the mass of 0.1 mol Na₂CO₃:

mass of 0.1 mol Na₂CO₃ = n x molar mass = (0.1 mol)(105.98 g/mol) = 10.598 g.

∵ The percent yield = [actual yield]/[theoretical yield] x 100.

∴ 81.0 % = [actual yield]/[10.598 g] x 100.

∴ [actual yield] = (81.0 %)(10.598)/(100) = 8.58 g.

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Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

Explanation:

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First of all one should look up the reduction potentials for the species envolved:

$Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

$Fe^{3+} + e$ → $Fe^{2+}$ E°red=0.771V

2) Redox pair

Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).

1) $Ce^{4+}$ reduces while $Fe^{2+}$ oxidates

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The cell potential can also be calculated as the cathode potencial minus the anode potential:

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ΔG°=-nFE°=-1*96500*0.839

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E=E°- $\frac{RT}{nF} Ln Q$

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