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

Feb 10, 2015

I assume you meant how many moles of bismuth are needed to react with $\text{1.58 moles}$ of oxygen, right?

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

$4 B {i}_{\left(s\right)} + 3 {O}_{2 \left(g\right)} \to 2 B {i}_{2} {O}_{3 \left(s\right)}$

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 $\text{4:3}$ mole ratio between bismuth and oxygen. Since you were told that you have $\text{1.58 moles}$ of oxygen, use the mole ratio to determine how many moles of bismuth you would need for the reaction

$\text{1.58 moles oxygen" * ("4 moles bismuth")/("3 moles oxygen") = "2.11 moles Bi}$

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

$\text{2.11 moles Bi" * (6.022 * 10^(23)"atoms of Bi")/("1 mole Bi") = 1.27 * 10^(24)"atoms}$