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Answer:
d. 127 g/mol.
Explanation:
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In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:
Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:
It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.
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Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of butane
b) moles of oxygen
According to stoichiometry :
2 moles of butane require 13 moles of
Thus 0.09 moles of butane will require = of
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.