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puteri [66]
2 years ago
10

Calculate your % yield of co2 in the reaction based on the grams of nahco3 being the limiting reagent in the reaction.

Chemistry
1 answer:
Lynna [10]2 years ago
3 0

The percent yield of CO₂ is 93.3%.

<h3>What is the percent yield of CO₂?</h3>

The percent yield of a substance is given as follows:

  • Percent yield = actual yield/theoretical yield * 100 %

The equation of the reaction is used to determine the theoretical yield.

  • NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂

Moe ratio of sodium bicarbonate and CO₂ is 1 : 1.

Given that sodium bicarbonate is the limiting reactant, the theoretical yield of CO₂ will be:

Moles of NaHCO₃ reacting = 2.01/84 = 0.0239 moles

Theoretical yield of CO₂ = 0.0239 moles * 22.4L/mol = 0.536 L

Actual yield = 0.50 L

Percent yield = 0.50/0.536 * 100%

Percent yield = 93.3%

In conclusion, the percent yield is the ratio of the actual yield and theoretical yield.

<em>Note that the complete question is given below:</em>

<em>Calculate your % yield of co2 in the reaction based on the grams of nahco3 being the limiting reagent in the reaction between 2.01 g of sodium bicarbonate and 24.6 mL of 1.5 M acetic acid? They produce 0.50 L of at s.t.p.</em>

<em />

Learn mores about percent yield at: brainly.com/question/8638404

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For the following reactions, write a balanced equation using half-reactions and calculate the voltage to be expected.
Iteru [2.4K]

a) 2Na(s) +2H_2O(I)→ 2Na^+(aq) + OH^-(aq)+ H_2(g)

b) 2Ag (s) +2H^(aq) → 2 Ag^+ (aq) +H_2(g)

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

a)

2Na(s) +2H_2O(I)→ 2Na^+(aq) + OH^-(aq)+ H_2(g)

E^0 cell = E^0 (reduction)  - E^0 (oxidation)

= E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)

= -0.83 - (-2.71) =1.88V

b)

2Ag (s) +2H^(aq) → 2 Ag^+ (aq) +H_2(g)

E^0cell= E^0(H^+/H_2) -E^0(Ag^+/Ag)

E^0cell=-0. - (0.8) =-0.8V

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8 0
2 years ago
Which organism is able to cause an infection
spin [16.1K]

Answer: The answer is bacteria. The correct answer is The letter B or the seconed one!

8 0
3 years ago
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Natural gas (CH4) has a molar mass of 16.0 g/mole. You started out the day with a tank containing 200.0 g of natural gas. At the
hodyreva [135]

Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas.  So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}

<u><em>amount of moles at the beginning= 12.5 moles</em></u>

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}

<u><em>amount of moles at the end= 1.875 moles</em></u>

The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.

moles used= amount of moles at the beginning - amount of moles at the end

moles used= 12.5 moles - 1.875 moles

<u><em>moles used= 10.625 moles</em></u>

<u><em /></u>

Finally, the moles of gas used are 10.625 moles.

Learn more about molar mass:

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3 0
2 years ago
The escape of gas through a small hole in a container
Doss [256]
<span>The escape of gas through a small hole in a container is called effusion. This phenomenon happens when the diameter of the hole is small enough compared to the mean free path of the gas particles. This is governed by Graham's Law which states that the rate of effusion is inversely proportional to the molecular weight of the gas.</span>
7 0
3 years ago
Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO
zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons  4NOBr ( g )

The suitable equilibrium constant turns out:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Or in terms of the initial equilibrium constant:

K_2=K_1^2

Since the second reaction is a doubled version of the first one.

Best regards.

5 0
3 years ago
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