Question

What mass of gallium oxide, Ga2O3, can be prepared from 2.14 x 1024 molecules of oxygen? 4Ga+3O2⟶2Ga2O3

Answer 1

According to our question, we need to find the mass of Ga2O3, that can be prepared from 2.14 x 10^24 molecules of O2.

Step 1) It says: mol O2/a O2 molecules. To find "a", we need to remember this:

1 mol of any substance = 6.022x10^23 formula units (atoms, molecules, ions, etc), in this case, we have molecules of O2

So, 1 mol O2 = 6.022x10^23 O2 molecules. Therefore:

a = 6.022x10^23

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Step 2) b mol Ga2O3/ c mol O2

To know b and c, you must pay attention to their relation in the reaction, and this is (balanced reaction always) 2 moles Ga2O3 / 3 moles O2

So, b = 2 and c = 3

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Step 3) d g Ga2O3 / mol Ga2O3

For d, you need the molar mass of Ga2O3 (use the periodic table),

187.4 g/mol

So, 1 mol of Ga2O3 = 187.4 g, then 187.4 g Ga2O3/ mol Ga2O3

Therefore, d = 187.4 g

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At last, we gather everything, and do the math:

`2.14x10^(24)\text{molec x }\frac{mol\text{ O2}}{6.022x10^(23)O2\text{molec}}x\frac{2\text{ mol Ga2O3}}{3\text{ mol O2}}x\frac{187.4\text{ g Ga2O3}}{\text{mol }Ga2O3}=\text{ 444.0 g Ga2O3}`

Answer: e = 444.0 g Ga2O3