In fractional distillation, liquid can be seen running from the bottom of the distillation column back into the distilling flask

Answer:

The percent yield of Br₂ in this reaction amounts to 96.15%

Explanation:

The reaction's balanced stoichiometric equation is:

2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂

To calculate the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

To determine the theoretical yield:

5.29 g of NaBr reacts with an excess of chlorine; therefore, NaBr is the limiting reagent, controlling the possible yield of products.

We convert 5.29 g of NaBr to moles.

Number of moles = (Mass)/(Molar mass)

Molar Mass of NaBr = 102.894 g/mol

Number of moles = (5.29/102.894) = 0.0514121329 = 0.05141 mole

According to the stoichiometry of the reaction:

2 moles of NaBr yield 1 mole of Br₂

Thus, 0.05141 mole of NaBr will produce (0.05141×1/2) mole of Br₂, which is 0.0257 mole of Br₂

Theoretical yield = Expected mass of Br₂ from the reaction

= (Number of moles) × (Molar mass)

Molar mass of Br₂ = 159.808 g/mol

Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g

Calculating the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

Actual yield = 3.95 g

Theoretical yield = 4.108 g

Percent yield = 100% × (3.95/4.108) = 96.15%

Hope this is helpful!!!

The equation representing the reaction between sodium bicarbonate and hydrochloric acid is as follows:

The substances and combine in a 1:1 ratio. Therefore, we calculate the quantity of sodium bicarbonate and its molar mass to determine the moles formed.

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We also recognize that the stoichiometric proportions are 1:1:1:1:1, which leads to the conclusion that the moles of equal 24.977 moles.

Next, we apply the ideal gas equation , where P denotes pressure, V refers to volume, R is the gas constant, and T represents the temperature in kelvins. We rearrange to solve for V

The final answer should be expressed in liters, , hence