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sammy [17]
3 years ago
5

Give the percent yield when 162.8 g of CO2 are formed from the reaction of excess amount of

Chemistry
2 answers:
Zepler [3.9K]3 years ago
6 0

Answer:

84.86%

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction between C8H18 and O2 to produce CO2. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Step 2:

Now, let us calculate the mass of O2 that reacted and the mass of CO2 produced from the balanced equation above. This is illustrated below:

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 25 x 32 = 800g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Therefore the mass of O2 that reacted from the balanced equation is 800g

The mass of CO2 produced from the balanced equation is 704g

Step 3:

Determination of the theoretical yield of CO2. This is illustrated below:

From the balanced equation above,

800g of O2 reacted to produced 704g of CO2.

Therefore, 218g of O2 will react to produce = (218 x 704)/800 = 191.84g of CO2.

Therefore, the theoretical yield of CO2 is 191.84g

Step 4:

Determination of the percentage yield of CO2. This is illustrated below:

Actual yield = 162.8g

Theoretical yield = 191.84g

Percentage yield =?

Percentage yield = Actual yieldm/Theoretical yield x100

Percentage yield = 162.8/191.84 x100

Percentage yield = 84.86%

Therefore, the percentage yield of CO2 is 84.86%

zhannawk [14.2K]3 years ago
5 0

Answer:

The percent yield of the reactions is  84.8 %

Explanation:

Step 1: data given

Mass of CO2 formed = 162.8 grams

Mass of O2 = 218.0 grams

C8H18 is in excess

Molar mass O2 = 32.0 g/mol

Molar mass of CO2 = 44.01 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles O2

Moles O2 = 218.0 grams / 32.0 g/mol

Moles O2 = 6.8125 moles

Step 4: Calculate moles CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.8125 moles O2 we'll have 16/25 * 6.8125 = 4.36 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 4.36 moles ¨¨ 44.01 g/mol

Mass CO2 = 191.9 grams

Step 6: Calculate the percent yield

Percent yield = (actual mass / theoretical mass) * 100 %

Percent yield = (162.8 grams / 191.9 grams ) * 100 %

Percent yield = 84.8 %

The percent yield of the reactions is  84.8 %

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3 years ago
Suppose you mix one mole of sulfuric acid (H2SO4) with 1 mole of sodium hydroxide(NaOH).
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<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O,</em>

It is clear that every 1.0 mol of H₂SO₄ needs 2 mol of NaOH to be neutralized completely.

<em>So, when you mix one mole of sulfuric acid with 1 mole of sodium hydroxide, there will be an excess of sulfuric acid.</em>

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<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

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Molarity of MnSO_4 = 0.796 M

Volume of MnSO_4 = 161 mL = 0.161 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

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By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = \frac{1}{1}\times 0.13=0.13mol of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L

Converting this into milliliters, we use the conversion factor:

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So, 0.183L=0.183\times 1000=183mL

Hence, the volume of barium hydroxide is 183 mL.

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