Question

Which is the limiting reactant when 5.00 g of H₂ and 10 g of O₂ react to form water?

Short answer

O₂ is the limiting reactant

Step-by-step solution

Calculating moles of both of them

The balanced equation is as follows

\(2{H_2} + {O_2} \to 2{H_2}O\)

So, moles of H₂

\(\begin{array}{l} = \frac{{{\rm{Given}}\,{\rm{mass}}}}{{{\rm{Molecular}}\,{\rm{weight}}}}\\ = \frac{{5g}}{{2g}}\\ = 2.5\,{\rm{moles}}\end{array}\)

So, moles of O₂

\(\begin{array}{l} = \frac{{{\rm{Given}}\,{\rm{mass}}}}{{{\rm{Molecular}}\,{\rm{weight}}}}\\ = \frac{{10g}}{{32g}}\\ = 0.3125\,{\rm{moles}}\end{array}\)

Determine the limiting reactant

From the balanced equation, we can see that 2 mol of H₂ produces 2 mol of H₂O

So, 2.5 mol of H₂ will produce = 2.5 x 2/2= 2.5 moles of H₂O

1 mol of O₂ produces 2 mol of H₂O

0.1325 mol will produce = 0.3125 x 2 /1= 0.625 moles of H₂O

As the amount of O₂ is much smaller than H₂. Hence the limiting reactant is O₂.