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Lena [83]
3 years ago
10

How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?

Chemistry
1 answer:
Lana71 [14]3 years ago
8 0
Hey...
Use the molarity formula
M=moles/L and then convert to grams

0.07268*0.15=moles
<span>0.010902 mol
</span>Pb(NO3)2
1 mole=331.22g
0.010902 moles=

3.61 g


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A solution of methanol and water has a mole fraction of water of 0.312 and a total vapor pressure of 21 torr at 39.9 degrees C.
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Chem quiz please help
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The theoretical and percentage yield for the reaction are:

  • The theoretical yield is 21 g
  • The percentage yield is 119%

<h3>Balanced equation </h3>

CH₄ + 2O₂ —> CO₂ + 2H₂O

Molar mass of CH₄ = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂ to produce 44 g of CO₂

<h3>How to determine the limiting reactant</h3>

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂

Therefore,

20 g of CH₄ will react with = (20 × 64 ) / 16 = 80 g of O₂

From the above calculation, a higher mass (i.e 80 g) of O₂ than what was given (i.e 30 g) is needed to react completely with 20 g of CH₄.

Therefore, O₂ is the limiting reactant

<h3>How to determine the theoretical yield of CO₂</h3>

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂

Therefore,

30g of O₂ will react to produce = (30 × 44) / 64 = 21 g of CO₂

<h3>How to determine the percentage yield </h3>
  • Actual yield of CO₂ = 25 g
  • Theoretical yield of CO₂ = 21 g
  • Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (25 / 21) ×100

Percentage yield = 119%

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