Answered

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How much of the excess reagant in Problem 1 is left over?

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Answer:

4 Al + 3 O2 → 2 Al2O3

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al

(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =

10.1 g O2 left over

Explanation:

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