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Katarina [22]
3 years ago
6

Lead(II) oxide from an ore can be reduced to elemental lead by heating in a furnace with carbon. Calculate the expected yield of

lead if 57.0 kg of lead oxide is heated with 57.0 kg of carbon. Expected yield
Chemistry
1 answer:
Lelu [443]3 years ago
6 0

Answer:

53kg is the expected yield of lead

Explanation:

Firstly, in order to solve this question, we need to write the equation of reaction correctly. This is as follows:

PbO(s) + C(s) ---> Pb(l) + CO(g)

We proceed from here. We should get the limiting reactant but this can only be obtained by getting the number of moles of each reactant present.

The formula to use across all boards is that the number of moles is the mass of each of the reactant divided by the molar mass of each of the reactant.

For PBO, mass is 57kg = 57000g

Molar mass of PBO = 223.20g/mol

The number of moles is thus 57,000/223.2 = 255.37 moles

For carbon, mass is also 57kg = 57000g

Molar mass is 12g/mol

Number of moles of carbon = 57000/12 = 4750 moles

From the number of moles, we can see that the number of moles of Carbon is greater than that of PbO. This means that PbO is the limiting reagent.

Hence we use it to calculate percentage yield.

The number of moles of lead formed is the same of number of moles of lead oxide = 255.37 since we have mole ratio of 1 to 1

The molar mass of lead is 207.20g/mol

The mass of lead formed is = moles of lead formed * molar mass of lead = 207.20 * 255.37 = 52,912g which is approximately 53kg

Hence the expected yield is 53kg

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(a) One form of the Clausius-Clapeyron equation is

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Solving for ΔHv:

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(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

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