What mass of Fe can be produced by the reaction of 75.0 g CO with excess Fe2O3 according to the equation: Fe2O3(s) + CO(g) → Fe(
2 answers:
Here we have to get the mass of the iron (Fe) produces in the reaction between ferric oxide (Fe₂O₃) and carbon monooxide (CO) to produce carbon dioxide and iron.
The amount of Fe produced in the reaction is B. 99.7g.
The balanced form of the reaction given is: Fe₂O₃ (s) + 3CO = 2Fe + 3CO₂.
Now as per the reaction 3 moles of CO produce 2 moles of Fe.
The molar mass of CO and Fe are 28.01g/mol and 55.845g/mol.
Thus (3×28.01) 84.03g of CO produces 111.69g of Fe.
Hence, 75.0g of CO produces ×75.0 = 99.687g of Fe or 99.7g of Fe.
A. If in the reaction 49.8g Fe is produced in the reaction then ×49.8 = 37.467g of CO should react. Thus the option is wrong.
C. To produce 224g of Fe ×224 = 168.526g of CO will consumed. Thus it is wrong statement.
D. To generate 299g of Fe ×299 = 224.952g of CO will consumed. Thus it is wrong statement.
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Answer:
Explanation:
We usually approximate the density of water to about at room temperature. In terms of the precise density of water, this is not the case, however, as density is temperature-dependent.
The density of water decreases with an increase in temperature after the peak point of its density. The same trend might be spotted if the temperature of water is decreased from the peak point.
This peak point at which the density of water has the greatest value is usually approximated to about . For your information, I'm attaching the graph illustrating the function of the density of water against temperature where you could clearly indicate the maximum point.
To a higher precision, the density of water has a maximum value at , and the density at this point is exactly
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