What is the percent composition of Sb2F3.What is the percent composition of F in this molecule? Round the answer to five signifi
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Answer:
Explanation:
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In this case, a percent composition is computed by considering the atomic mass of each element in the compound as well as the number of atoms and the molar mass of the compound. Thus, Sb2F3 has a molar mass of 300.52 g/mol, antimony has an atomic mass of 121.76 g/mol and fluorine 19.0 g/mol, therefore, the percent composition of fluorine which has three atoms with five significant figures is:
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The limiting reagent will be Al
<h3>What are limiting reagents?</h3>They are reagents that limit the quantity of products that are formed in reactions.
From the equation of the reaction:
The mole ratio of Al to O2 is 4:3.
With 2 moles of Al and 2 moles of O2, Al becomes limiting while O2 is in excess.
With 2 moles of O2, the amount of Al required should be:
2 x 4/3 = 2.67 moles.
With 2 moles of Al, the amount of O2 required should be:
2 x 3/4 = 1.5 moles
Thus, O2 is in excess by 0.5 moles.
More on limiting reagents can be found here: brainly.com/question/11848702
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