<h3><u>Answer</u>;</h3>
Actual yield = 46.44 g
<h3><u>Explanation;</u></h3>
1 mole of water = 18 g/mol
Therefore;
The experimental yield = 2.58 moles
equivalent to ; 2.58 × 18 = 46.44 g
The theoretical value is 47 g
Percentage yield = 46.44/47 × 100%
= 98.8%
The questions asks for actual yield = 46.44 g
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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Answer:
1528.3L
Explanation:
To solve this problem we should know this formula:
V₁ / T₁ = V₂ / T₂
We must convert the values of T° to Absolute T° (T° in K)
21°C + 273 = 294K
70°C + 273 = 343K
Now we can replace the data
1310L / 294K = V₂ / 343K
V₂ = (1310L / 294K) . 343K → 1528.3L
If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too
You just have to pass your final with a grade above 75
