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3 years ago

What is the mass of an atom that has 4 protons, 5 neutrons, and 4 electrons?

Chemistry

2 answers:

Mazyrski [523]3 years ago

3 0

The answer would be 9

WITCHER [35]3 years ago

3 0

To get the mass, you would add the 4 protons and 5 neutrons so it would be 9.
(The electrons weigh practically nothing so you just ignore them.)

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3 years ago

Read 2 more answers

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

STALIN [3.7K]

Answer: The theoretical yield of barium sulfate is 50.9 grams

Explanation:

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

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

For sodium sulfate:

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

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

For barium chloride:

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

7 0

3 years ago