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Answer 1 · 2017-05-15T17:32:56

$0.246 {mol of H}_{2}$ $0.246"mol of H"_2$

To use Dimensional Analysis (the Factor-Label Method), you begin by writing the given quantity over 1:

$\frac{0.123 {mol of O}_{2}}{1}$ $(0.123"mol of O"_2)/1$

We observe that two moles of hydrogen are need for 1 mole of oxygen, therefore we multiply by that conversion factor:

$\frac{0.123 {mol of O}_{2}}{1} \frac{2 {mol of H}_{2}}{1 {mol of O}_{2}}$ $(0.123"mol of O"_2)/1(2"mol of H"_2)/(1"mol of O"_2)$

Please observe how the units cancel:

$(0.123cancel("mol of O"_2))/1(2"mol of H"_2)/(1cancel("mol of O"_2))$

We may stop and perform the multiplication, because the only unit that remain are the desired units:

$(0.123cancel("mol of O"_2))/1(2"mol of H"_2)/(1cancel("mol of O"_2)) = 0.246"mol of H"_2$