Answer:
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Explanation:
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Answer: a. 7.31 g
b. 20.4 %
Explanation:
To calculate the moles :
According to stoichiometry :
3 moles of require = 1 mole of
Thus 0.645 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 3 moles of give = 2 moles of
Thus 0.645 moles of give = of
Mass of
Thus theoretical yield for this reaction under the given conditions is 7.31 g.
b)
The percent yield for this reaction under the given conditions is 20.4 %
Answer:
Explanation:
To convert from moles to mass, the molar mass must be used.
First, write the chemical formula for carbon monoxide. Since the carbon (C) comes first without a prefix, there is 1 carbon atom. The prefix mono- before oxide means 1, so there is also 1 oxygen (O) atom. The formula is CO.
Next, look up their molar masses on the Periodic Table.
- C: 12.011 g/mol
- O: 15.999 g/mol
Since there is 1 atom of each, the molar masses can be added.
- CO: 12.011 g/mol + 15.999 g/mol = 28.01 g/mol
Use this molar mass as a ratio.
Multiply by the given number of moles:5.55
The moles of carbon monoxide cancel.
Multiply.
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, it is the ones place. The 4 in the tenths place tells us to leave the 5.
5.55 moles of carbon monoxide is about 155 grams.