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mario62 [17]
3 years ago
6

If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reac

tion below: Na2SO4 (aq) + BaCl2 (aq) → BaSO4 (s) + 2NaCl (aq) What is the theoretical yield of barium sulfate (BaSO4) in grams?
Chemistry
1 answer:
STALIN [3.7K]3 years ago
7 0

<u>Answer:</u> The theoretical yield of barium sulfate is 50.9 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For sodium sulfate:</u>

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol

  • <u>For barium chloride:</u>

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = \frac{1}{1}\times 0.228=0.228mol of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = \frac{1}{1}\times 0.228=0.228moles of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g

Hence, the theoretical yield of barium sulfate is 50.9 grams

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A mixture of sodium bicarbonate and ammonium bicarbonate is 75.9% bicarbonate by mass. what is the mass percent of sodium bicarb
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Justification:

The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:

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.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.

Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.

With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.

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