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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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PbSO₄ partially dissociates in water. the balanced equation is;
                    
                       PbSO₄(s) ⇄  Pb²⁺(aq) + SO₄²⁻(aq)
Initial                                     -                -
Change             -X               +X           +X
Equilibrium                           X              X

Ksp           =    [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸  =    X * X
1.6 x 10⁻⁸  =    X²
          X    =   1.3 x 10⁻⁴ M
      
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M. 
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8 0
3 years ago
Given:
bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

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3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

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mass of oxygen per gram sulfur = 3.45 g / 3.46 g
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What is the mass of oxygen in 10.0 g of water?
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