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Lapatulllka [165]
3 years ago
7

Exercise 7.46 We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat = 0

.1291 J⋅g−1⋅K−1 ) with mass of 10.5 g is heated to 78.3 ∘C and immersed in 50 g of 23.7 ∘C water in a constant-pressure calorimeter. The final temperature of the water is 31.0 ∘C. You may want to reference ( page ) Section 7.1 while completing this problem.
Chemistry
1 answer:
RUDIKE [14]3 years ago
6 0
So, the question would would probably ask if the gold ring is indeed pure gold. Let's calculate the specific heat capacity of the calorimeter. If this is equal to the given specific heat for pure gold, then the the gold ring is pure gold.

Qwater = mCdT = (50 g)(4.18 J/gC)(31 - 23.7) = 1525.7 J
Through conservation of energy,
Qwater = Qcalorimeter = mCdT = 1525.7
1525.7 = (10.5)(C)(78.3 - 31)
Solving for C,
C = 3.072 J/gC

Since the specific heat of the calorimeter is not equal to that of the pure gold (0.1291 J/gC), then the gold ring is not pure.
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A reaction vessel for synthesizing ammonia by reacting nitrogen and hydrogen is charged with 5.79 kg of h2 and excess n2. a tota
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82.08 %

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  • <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100. </em>

Actual yield = 26.80 kg.

  • <em><u>To get the calculated yield: </u></em>
  • The balanced equation of reacting N2 with H2 to produce NH3 is:

N₂ + 3H₂ → 2NH₃

  • It is clear that 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
  • N₂ is present in excess and H₂ is the limiting reactant.
  • We need to convert the mass of H₂ added (5.79 kg) to moles using the relation:

n = mass /molar mass = (5790 g) / (2.01 g/mol) = 2880.6 mol.

  • We can get the no. of moles of NH₃ produced.

<u><em>Using cross multiplication: </em></u>

3.0 mole of H₂ produce → 2.0 moles off NH₃, from the stichiometry.

2880.6 mol of H₂ produces → ??? moles of NH₃.

  • The no. of moles of NH₃ produced = (2880.6 mol)(2.0 mol) / (3.0 mol) = 1920.4 mol.
  • We can know get the calculated yield of NH₃ = no. of moles x molar mass = (1920.4 mol) (17.00 g/mol) = 32646.76 g ≅ 32.65 kg.

∴ <em>The percent yield of the reaction = [(actual yield)/(calculated yield)] x 100 = </em>[(26.8 kg) / (32.65 kg)] x 100 <em>= 82.08 %.</em>

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Have a nice day!

7 0
4 years ago
Read 2 more answers
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