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vlada-n [284]
1 year ago
8

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

eaction? (molar mass of K2O = 94.20 g/mol)4 K(s) + O2(g) → 2 K20(s)a. 87.54%b. 95%c. 94.85%d. 88%
Chemistry
1 answer:
Mumz [18]1 year ago
3 0

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

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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Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂  → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • CuCl₂: 1 mole
  • CuCO₃: 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

  • Na₂CO₃: 129 g/mole
  • CuCl₂: 134.45 g/mole
  • CuCO₃: 123.55 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Na₂CO₃: 1 mole ×129 g/mole= 129 grams
  • CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
  • CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
  • NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

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A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
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Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

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