Question

A mixture of \(80.0 \mathrm{~mL}\) of hydrogen and \(60.0 \mathrm{~mL}\) of oxygen is ignited by a spark to form water. (a) Does any gas remain unreacted? Which one, \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ? (b) What volume of which gas (if any) remains unreacted? (Assume the same conditions before and after the reaction.)

Short answer

Yes, 20.0 mL of oxygen remains unreacted.

Step-by-step solution

- Write the balanced chemical equation

The balanced chemical equation for the combustion of hydrogen in oxygen to form water is:\[ 2H_2 + O_2 \rightarrow 2H_2O \]

- Determine the stoichiometric ratio

From the balanced equation, 2 volumes of hydrogen react with 1 volume of oxygen. This means the stoichiometric ratio is 2:1 for hydrogen to oxygen.

- Calculate the needed oxygen for given hydrogen

Given 80.0 mL of hydrogen, the volume of oxygen required can be calculated using the stoichiometric ratio:\[ \text{Volume of } O_2 = \frac{80.0 \text{ mL } H_2}{2} = 40.0 \text{ mL} \]

- Determine the remaining oxygen

Since initially 60.0 mL of oxygen is provided and only 40.0 mL is required to react with 80.0 mL of hydrogen, the volume of unreacted oxygen is:\[ 60.0 \text{ mL} - 40.0 \text{ mL} = 20.0 \text{ mL} \]

Key concepts

balanced chemical equation

In chemistry, it's important to have a balanced chemical equation to understand how substances react with each other. A balanced equation has the same number of each type of atom on both sides of the reaction.

For example, in the combustion reaction of hydrogen and oxygen to form water, the balanced equation is: \[ 2H_2 + O_2 \rightarrow 2H_2O \].

This equation tells us that two molecules of hydrogen gas (\(H_2\)) react with one molecule of oxygen gas (\(O_2\)) to form two molecules of water (\(H_2O\)).
Balance is achieved here because we have:

• 4 hydrogen atoms on the reactant side (2 molecules of H2 x 2 atoms per molecule)
• 4 hydrogen atoms on the product side (2 molecules of H2O x 2 atoms per molecule)
• 2 oxygen atoms on the reactant side (1 molecule of O2 x 2 atoms per molecule)
• 2 oxygen atoms on the product side (2 molecules of H2O x 1 atom per molecule)

This balance allows us to understand how much of each reactant we need and how much product will be formed.

Balancing chemical equations is crucial for stoichiometry, which we will discuss next.

stoichiometric ratio

The stoichiometric ratio in a chemical reaction tells us how the quantity of reactants compares to the quantity of products.

For the reaction between hydrogen and oxygen, the stoichiometric ratio is 2:1. This means it requires 2 volumes of hydrogen gas to react with 1 volume of oxygen gas.

This ratio can be determined from the balanced chemical equation: \[ 2H_2 + O_2 \rightarrow 2H_2O \].
From here, we see that 2 volumes of \(H_2\) react with 1 volume of \(O_2\), producing 2 volumes of \(H_2O\).

Let's use this information to solve a practical problem. If we have 80.0 mL of hydrogen gas, we can use the stoichiometric ratio to find out how much oxygen gas is needed.

• According to the stoichiometric ratio, for every 2 volumes of \(H_2\), 1 volume of \(O_2\) is needed.
• Therefore, if we have 80.0 mL of \(H_2\), we need half of that volume in \(O_2\), which is 40.0 mL.

Using stoichiometric ratios, we can precisely calculate the amounts of reactants and products involved in any chemical reaction.
Proficiency in this skill is extremely helpful in chemistry.

combustion reaction

A combustion reaction occurs when a substance reacts with oxygen, releasing energy in the form of light or heat.

One of the most common examples of a combustion reaction is the burning of hydrogen gas in oxygen to form water.

• In this case, the balanced chemical equation is: \[ 2H_2 + O_2 \rightarrow 2H_2O \]

To delve deeper, let’s address our exercise. We started with 80.0 mL of hydrogen and 60.0 mL of oxygen.

• Given the stoichiometric ratio(2:1), 80.0 mL \(H_2\) reacts with 40.0 mL \(O_2\)
• This leaves 20.0 mL of \(O_2\) unreacted, because 60.0 mL was initially available

Such combustion reactions are crucial in various fields like energy production and automotive engines.
Mastery of these concepts aids in understanding broader scientific phenomena.