How many mole of Bi atoms are needed to combine with 1.58 O atoms to make
Bi2O3?

Question

Noah Aldrich · Accepted Answer

I assume you meant how many moles of bismuth are needed to react with #"1.58 moles"# of oxygen, right?

Well, here's the balanced chemical equation for your reaction

#4Bi_((s)) + 3O_(2(g)) -> 2Bi_2O_(3(s))#

Notice that the balanced chemical equation has 4 moles of bismuth reacting with 3 moles of oxygen to form 2 moles of bismuth (III) oxide.

What that means is that you have a #"4:3"# mole ratio between bismuth and oxygen. Since you were told that you have #"1.58 moles"# of oxygen, use the mole ratio to determine how many moles of bismuth you would need for the reaction

#"1.58 moles oxygen" * ("4 moles bismuth")/("3 moles oxygen") = "2.11 moles Bi"#

If you're interested in how many actual atoms of bismuth you'd need, use Avogadro's number to go from moles to atoms

#"2.11 moles Bi" * (6.022 * 10^(23)"atoms of Bi")/("1 mole Bi") = 1.27 * 10^(24)"atoms"#

Cameron Andrews · Answer

undefined To form Bi₂O₃, we need 3 oxygen atoms per Bi atom. Therefore, to combine with 1.58 oxygen atoms, we would need $ \frac{1.58}{3} $ Bi atoms. This equals approximately 0.53 Bi atoms.

How many mole of Bi atoms are needed to combine with 1.58 O atoms to make Bi2O3?