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The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g
We'll begin by writing the balanced equation for the reaction. This is given below:
2Fe₂O₃ -> 4Fe + 3O₂
- Molar mass of Fe₂O₃ = 159.7 g/mol
- Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
- Molar mass of Fe = 55.85 g/mol
- Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
<h3>How to determine the mass of iron, Fe produced</h3>
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
Therefore,
21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe
Thus, 15.04 g of Fe were produced.
Learn more about stoichiometry:
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I believe it’s the third option
Chemically combined to make a new pure substance
Answer:
PV=nRt
Therefore n(number of moles)=PV/RT
=>(0.49×3.80)/(0.08206×320)
Therefore Number of moles is = 0.071mols
Explanation: By using the Real gas equation..
PV=NRT .
We can solve for the number of moles of Ar by making N the subject..
Always make sure you pressure is In atm, your Volume is in Litres and temperature in degree Kelvin.
Also Recall the universal gas constant R used in this type of questions which is 0.08206.
Hence l, by making N the subject we get our answer as
