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erica [24]
3 years ago
15

Consider the following reaction: NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2 How many mol of CO2 would be produced from the complete

reaction of 25 mL of 0.833 mol/L HC3H3O2 with excess NaHCO3 ?
Chemistry
2 answers:
Assoli18 [71]3 years ago
6 0

Answer:

0.208mole of CO2

Explanation:

First, let us calculate the number of mole of HC3H3O2 present.

Molarity of HC3H3O2 = 0.833 mol/L

Volume = 25 mL = 25/100 = 0.25L

Mole =?

Mole = Molarity x Volume

Mole = 0.833 x 0.25

Mole of HC3H3O2 = 0.208mole

Now, we can easily find the number of mole of CO2 produce by doing the following:

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2

From the equation,

1mole of HC2H3O2 produced 1 mole of CO2.

Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2

scoray [572]3 years ago
3 0

Answer:

For 0.0208 moles CH2H3O2 we'll have 0.0208 moles CO2 produced

Explanation:

Step 1: Data given

Volume of HC2H3O2 = 25 mL = 0.025 L

Concentration of HC2H3O2 = 0.833 mol / L

NaHCO3 is in excess

Step 2: The balanced equation

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2

For 1 mol NaHCO3 we need 1 mol HC2H3O2 to produce 1 mol NaC2H3O2, 1 mol H2O and 1mol CO2

Step 3: Calculate moles HC2H3O2

Moles HC2H3O2 = concentration HC2H3O2 * volume solution

Moles HC2H3O2 = 0.833 mol/L * 0.025 L

Moles HC2H3O2 = 0.0208 moles

Step 4: calculate moles CO2

For 1 mol NaHCO3 we need 1 mol HC2H3O2 to produce 1 mol NaC2H3O2, 1 mol H2O and 1mol CO2

For 0.0208 moles CH2H3O2 we'll have 0.0208 moles CO2 produced

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9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r
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Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

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What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\  \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\  \\ Percent\text{ }yield=0.9484\times100\% \\  \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}

Therefore, the percent yield for the reaction is 95%.

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The bubbles mentioned in the question hints that our interest is the compounds in their gseous phase  (g).

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