Question 5.SP.21: Consider the reaction of hydrogen and oxygen to form water, ...
Consider the reaction of hydrogen and oxygen to form water, according to the balanced equation:
2H_{2}( g) + O_{2}( g) \longrightarrow 2H_{2}O( l).
Identify the limiting reactant in the molecular art, and draw a representation showing how much product is formed and what reactant molecules are left over.
Use the amount of each reactant and the mole ratio from the balanced equation to determine thelimiting reactant.Use the amount of limiting reactant to determine the amount of product formed.
[1] To determine how much of one reactant is needed to react with a second reactant, label one reactant as the original quantity and the second reactant as the unknown quantity.
• These labels are completely arbitrary. In this example we identify the number of molecules of H_{2} as the original quantity.
\begin{matrix} \text{4 molecules of } H_{2}& \text{? molecules of } O_{2} \\ \text{original quantity} & \text{unknown quantity} \end{matrix}[2] Write out the conversion factors that relate the number of moles (or molecules) of reactants.
• Use the coefficients in the balanced equation to write a mole–mole (or molecule–molecule) conversion factor for H_{2} \text{ and }O_{2}.
\begin{matrix} \frac{\text{2 molecules }H_{2}}{\text{1 molecule }O_{2}} & or & \frac{\text{1 molecule }O_{2}}{\text{2 molecules }H_{2}}\longleftarrow \text{Choose this conversion factor to cancel molecules of } H_{2}\end{matrix}
[3] Calculate the number of moles (or molecules) of the second reactant needed for complete reaction.
4 \cancel{\text{ molecules }} H_{2}\times \frac{1 \text{ molecules }O_{2}}{2 \cancel{\text{ molecules}} H_{2}} = 2 \text{ molecules of }O_{2} \text{needed}
[4] Analyze the two possible outcomes.
• If the amount present of the second reactant is less than what is needed, the second reactant is the limiting reactant.
• If the amount present of the second reactant is greater than what is needed, the second reactant is in excess.
In this reaction only two molecules of O_{2} are needed but three molecules of O_{2} are present, so O_{2} is present in excess and H_{2} is the limiting reactant. To draw the molecular art representing the product, we must know what happens to each reactant and how much product is formed.
• Since H_{2} is the limiting reactant, all of the H_{2} is consumed and none is left.
• Since two molecules of O_{2} react and three were present at the outset, one molecule of O_{2} is left over.
• Using the balanced equation, the number of molecules of product formed equals the number of molecules of the limiting reactant consumed. Thus, four molecules of H_{2} yield four molecules of H_{2}O .
| Summary | Reactants | Product | |||
| Equation | 2H_{2} | + | O_{2} | \longrightarrow | 2H_{2}O |
| \text{Initial quantities} | \text{4 molecules} | \text{3 molecules} | \text{ 0 molecules} | ||
| \text{Used or formed} | \text{−4 molecules} | \text{−2 molecules} | \text{+4 molecules} | ||
| \text{Molecules remaining} | \text{0 molecules} | \text{1 molecule} | \text{4 molecules} | ||
| \text{Limiting reactant} | \text{Excess reactant} | ||||