Answer:
Explanation:
Hello,
In this case, given the reaction, we can directly compute the theoretically yielded grams of carbon dioxide, considering the 2:2 molar ratio between carbon monoxide (molar mass = 28 g/mol) and carbon dioxide (molar mass = 44 g/mol) and the initial reacting grams of carbon monoxide in excess oxygen:
Thus, as only 8 g were actually yielded, we compute the percent yield:
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Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
mass = 0.508 g, Volume = 0.175 L
Temperature = (25 + 273) K = 298 K, P = 1 atm
As per the ideal gas law, PV = nRT.
where, n = no. of moles =
Hence, putting all the given values into the ideal gas equation as follows.
PV =
1 atm \times 0.175 L =
= 71.02 g
As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of is 70.8 g/mol or 71 g/mol (approx).
Thus, we can conclude that the gas is most likely chlorine.