What can you conclude about the mass of iron (III) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of yellow sulfur solid, S8(s)?

Answer : The mass of iron(III)sulfide is, 5.4288 g

Solution : Given,

Mass of iron, Fe = 3 g

Mass of sulfur, = 2.5 g

Molar mass of Fe = 56 g/mole

Molar mass of = 256 g/mole

Molar mass of iron(III)sulfide, = 208 g/mole

The balanced chemical reaction is,

First we have to calculate the moles of iron and sulfur.

From the balanced reaction, we conclude that

16 moles of Fe react with 3 moles of

0.054 moles of Fe react with moles of

Therefore, the excess reagent in this reaction is, Fe and limiting reagent is,

Now we have to calculate the moles of FeS.

As, 3 moles of gives 8 moles of

So, 0.0098 moles of gives moles of

The moles of = 0.0261 moles

Now we have to calculate the mass of .

Mass of = Moles of × Molar mass of

Mass of = 0.0261 g × 208 g/mole = 5.4288 g

Therefore, the mass of iron(III)sulfide is, 5.4288 g

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- "the choice that is not an indicator of a chemical change is "state of matter changes". more common than not, chemical reactions produce energy in the form of light or heat. along with energy, they also produce a new substance called the product that could be in any state of matter (solid, gas, or liquid)."

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.