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laiz [17]
2 years ago
12

If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

Chemistry
1 answer:
ANEK [815]2 years ago
4 0

Answer:

Approximately 64\%.

Explanation:

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

The actual yield of \rm O_2 was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g).

Note the ratio between the coefficient of \rm H_2O\, (g) and \rm O_2\, (g):

\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}.

This ratio will be useful for finding the theoretical yield of \rm O_2\, (g).

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

  • \rm H: 1.008.
  • \rm O: 15.999.

Calculate the formula mass of \rm H_2O and \rm O_2:

M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}.

M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}.

Calculate the number of moles of molecules in 9.8\; \rm g of \rm H_2O:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}.

Make use of the ratio \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2} to find the theoretical yield of \rm O_2 (in terms of number of moles of molecules.)

\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})}  \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}.

Calculate the mass of that approximately 0.271996\; \rm mol of \rm O_2 (theoretical yield.)

\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}.

That would correspond to the theoretical yield of \rm O_2 (in term of the mass of the product.)

Given that the actual yield is 5.6\; \rm g, calculate the percentage yield:

\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}.

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What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C
Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = \frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\

Now use the molar ratio to find how many moles of NaOH would be required to neutralize 5.5537*10^{-3} mol\\ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

         1           :            2                  (we get this from the balanced equation)

5.5537*10^{-3} mol\\ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L

5 0
3 years ago
Which of the following is a product of aerobic respiration?
sasho [114]

Answer:

The product of aerobic respiration is Carbon dioxide.

Explanation:

  • The process of breaking down glucose to produce energy and waste products is called respiration. Livings beings need respiration process to generate energy so  that they can survive.
  • The types of respiration are : Anaerobic and aerobic respiration.
  • Aerobic respiration takes place in presence of oxygen and produces large amount of energy.
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3 0
3 years ago
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Aqueous hydrobromic acid hbr will react with solid sodium hydroxide naoh to produce aqueous sodium bromide nabr and liquid water
Natalka [10]
 Balanced equation is

HBr + NaOH ----> NaBr + H2O

Using molar masses

80.912 g HBr reacts with  39.997 g of Naoh to give 18.007 g water

so 1 gram of NaOH reacts with 2.023 g of HBR   
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction

Mass  of water produced  =    (5.7 * 18.007 / 39.997  =  2.6 g to 2 sig figs
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A sample of gas has a volume of 4.4 liters when the pressure is 3.3 atm. What is the volume when the pressure is reduced to 2.2
murzikaleks [220]
6.6 liters would be your answer. 
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A sample of CO2 weighing 86.34g contains how many molecules?
irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

86.34 g CO₂

<u>Step 2: Identify Conversion</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

<u />86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.18141 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0
3 years ago
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