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nikitadnepr [17]
2 years ago
6

Chem quiz please help

Chemistry
1 answer:
GalinKa [24]2 years ago
4 0

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%

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

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A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the vo
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0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

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The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.

(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)

0.346 = (463*V2) / (777)

0.346*777 = (463*V2) / (777)*777

268.842 = 463*V2

268.842/463 = (463*V2)/463

V2 = 0.581

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How much force is needed to make a 60 kg object accelerate at a rate of 2.0 m/s2?
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Since the magnitude of force depends on the mass of the object and the acceleration by which it moves.

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g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.
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Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

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Divide by the smallest i.e 3.41

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Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

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