Question

How many milliliters of oxygen gas at STP are released from heating \(5.00 \mathrm{~g}\) of calcium chlorate? $$\mathrm{Ca}\left(\mathrm{ClO}_{3}\right)_{2}(s) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{~s})+3\mathrm{O}_{2}(g)$$

Short answer

From heating 5.00 g of calcium chlorate, 1620 mL of oxygen gas at STP is released.

Step-by-step solution

Determine the Molar Mass

Calculate the molar mass of calcium chlorate, \( \text{Ca}(\text{ClO}_3)_2 \). The molar masses are approximately: Ca = 40.08 g/mol, Cl = 35.45 g/mol, O = 16.00 g/mol. Thus, the molar mass of calcium chlorate is: \[\text{Ca}(\text{ClO}_3)_2 = 40.08 + 2\times(35.45 + 3\times16.00)\approx 206.98\, \text{g/mol}.\]

Calculate Moles of Calcium Chlorate

Calculate the number of moles of calcium chlorate using its mass. Given 5.00 g of \( \text{Ca}(\text{ClO}_3)_2 \), the moles are \( \frac{5.00\, \text{g}}{206.98\, \text{g/mol}} \approx 0.02415\, \text{mol}. \)

Determine Moles of Oxygen Gas Released

According to the balanced equation: \[\text{Ca}(\text{ClO}_3)_2 (s) \rightarrow \text{CaCl}_2 (s) + 3\text{O}_2 (g)\] 1 mole of \( \text{Ca}(\text{ClO}_3)_2 \) produces 3 moles of \( \text{O}_2 \). Thus, 0.02415 moles of \( \text{Ca}(\text{ClO}_3)_2 \) will produce \[0.02415 \times 3 = 0.07245\, \text{mol of } \text{O}_2.\]

Calculate Volume of Oxygen Gas at STP

Use the ideal gas law to find the volume at STP, where 1 mole = 22.4 L. Therefore, the volume of \( \text{O}_2 \) is \[0.07245\, \text{mol} \times 22.4\, \text{L/mol} = 1.62308\, \text{L}.\] Convert this to milliliters: \(1.62308\, \text{L} \times 1000 = 1623.08\, \text{mL}.\) Round to match significant figures based on the initial mass (3 significant figures): 1620 mL.

Key concepts

Molar Mass

Molar mass is like a bridge between the mass of a substance and how much of it we actually have in terms of molecules or atoms, which we call moles. To find the molar mass, you need to add up the weights of all the atoms in a molecule. For example, in the exercise, we calculated the molar mass of calcium chlorate, \(\text{Ca}(\text{ClO}_3)_2\). Each element in the compound has a specific atomic mass: calcium (Ca) is about 40.08 g/mol, chlorine (Cl) is approximately 35.45 g/mol, and oxygen (O) is about 16.00 g/mol.

To get the total molar mass of calcium chlorate:

• Start with calcium: 40.08 g/mol.
• Add chlorine: 2 \(\times\) 35.45 g/mol, since there are two chlorine atoms.
• Add oxygen: 2 \(\times\) 3 \(\times\) 16.00 g/mol, because each chlorine has three oxygens and there are two chlorines.

This gives a molar mass of approximately 206.98 g/mol.

Knowing the molar mass helps us convert from the mass of a substance to the number of moles, which is crucial for solving stoichiometry problems.

Balanced Chemical Equation

A balanced chemical equation ensures that we obey the law of conservation of mass, meaning the atoms on the reactant side equal the atoms on the product side. For instance, the given chemical reaction is:

\[\mathrm{Ca}(\mathrm{ClO}_3)_2 (s) \rightarrow \mathrm{CaCl}_2 (s) + 3\mathrm{O}_2 (g)\]

This equation is balanced. It shows that one molecule of calcium chlorate decomposes to form one molecule of calcium chloride and three molecules of oxygen gas.

Each side of the equation has:

• 1 calcium (Ca) atom
• 2 chlorine (Cl) atoms
• 6 oxygen (O) atoms

Balancing is crucial because it helps us determine the exact proportions of reactants and products, which is necessary to track during any chemical reaction.

These proportions allow us to calculate precisely how much of a product is formed when a certain amount of reactant is used, as seen in the exercise where the moles of oxygen were calculated from the given mass of calcium chlorate.

Ideal Gas Law

The ideal gas law is an important tool in chemistry for predicting how gases behave under different conditions. It's expressed as \( PV = nRT \), where:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant
• \( T \) is the temperature

However, for gases at standard temperature and pressure (STP), where the temperature is 0°C (273.15 K) and pressure is 1 atm, the volume of one mole of any gas is 22.4 liters. This simplifies many calculations!

In the exercise, after calculating the moles of oxygen produced from the reaction, we determine how much space it would take up using the ideal gas law under these STP conditions:

Since 1 mole of \(\text{O}_2\) occupies 22.4 L, multiplying the moles of \(\text{O}_2\) (0.07245 mol) gives us a volume of 1.62308 L. Converting this to milliliters gives us 1623.08 mL or roughly 1620 mL when rounded to match the significant figures of the starting data.

In conclusion, these calculated volumes and the simplification at STP are key to quickly determining gas behaviors in stoichiometric calculations.