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Karolina [17]
3 years ago
11

As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 15

0-mL Erlenmeyer flask. Is it large enough to contain the acid, the density of which is 1.83 g/mL?
Chemistry
1 answer:
Kamila [148]3 years ago
6 0

Answer:

Yes, the 150-mL Erlenmeyer will be large enough to contain the acid.

Explanation:

As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. Considering that the density of the phosphoric acid is 1.83 g/mL, we can find the volume occupied by the acid using the following expression.

density = mass / volume

volume = mass / density

volume = 225 g / (1.83 g/mL)

volume = 123 mL

The phosphoric acid occupies 123 mL so the 150-mL Erlenmeyer will be large enough to contain it.

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<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol

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BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = \frac{1}{1}\times 0.00286mol=0.00286mol of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

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