Question

Many metals react with oxygen gas to form the metal oxide. For example, calcium reacts as follows: $$ 2 \mathrm{Ca}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CaO}(s) $$ You wish to calculate the mass (g) of calcium oxide that can be prepared from \(4.20 \mathrm{~g}\) of \(\mathrm{Ca}\) and \(2.80 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (a) What amount (mol) of CaO can be produced from the given mass of Ca? (b) What amount (mol) of CaO can be produced from the given mass of \(\mathrm{O}_{2} ?\) (c) Which is the limiting reactant? (d) How many grams of CaO can be produced?

Short answer

5.87 grams of CaO can be produced.

Step-by-step solution

Molar Mass Calculation

Calculate the molar masses of calcium (Ca), oxygen (\text{O}_2), and calcium oxide (\text{CaO}).\(\text{Ca} = 40.08\ \text{g/mol}\)\(\text{O}_2 = 32.00\ \text{g/mol}\)\(\text{CaO} = 40.08\ \text{g/mol} + 16.00\ \text{g/mol} = 56.08\ \text{g/mol}\)

Calculate Moles of Calcium

Determine the amount of moles of calcium (\text{Ca}) using its mass and molar mass.\[ \text{moles of Ca} = \frac{4.20\ \text{g}}{40.08\ \text{g/mol}} = 0.1047\ \text{mol} \]

Calculate Moles of Oxygen

Determine the amount of moles of oxygen (\text{O}_2) using its mass and molar mass.\[ \text{moles of O}_2 = \frac{2.80\ \text{g}}{32.00\ \text{g/mol}} = 0.0875\ \text{mol} \]

Determine Moles of CaO from Ca

From the balanced chemical equation, \(2 \text{Ca}(s) + \text{O}_2(g) \rightarrow 2 \text{CaO}(s)\)For every 2 moles of Ca, 2 moles of CaO are produced. Therefore, moles of CaO from Ca:\[ \text{moles of CaO from Ca} = 0.1047\ \text{mol} \]

Determine Moles of CaO from O_2

From the balanced chemical equation, for every 1 mole of \text{O}_2, 2 moles of CaO are produced. Therefore, moles of CaO from O_2:\[ \text{moles of CaO from O}_2 = 0.0875\ \text{mol} \times 2 = 0.1750\ \text{mol} \]

Identify the Limiting Reactant

The limiting reactant is the one that produces the least amount of product. Comparing moles of CaO from Ca (0.1047 mol) and from O_2 (0.1750 mol), Calcium (Ca) is the limiting reactant.

Calculate Mass of CaO Produced

Use the moles of CaO produced by the limiting reactant (Ca) to find the mass of CaO. Use the formula: \(\text{mass} = \text{moles} \times \text{molar mass}\)\[ \text{mass of CaO} = 0.1047\ \text{mol} \times 56.08\ \text{g/mol} = 5.87\ \text{g} \]

Key concepts

Molar Mass Calculation

Understanding molar mass is crucial in stoichiometry. The molar mass of an element or compound is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). To find the molar mass of a compound, sum the molar masses of all the atoms in its formula, considering the number of each type of atom. For example, calcium oxide (CaO) consists of one calcium (Ca) and one oxygen (O) atom. The molar mass of Ca is 40.08 g/mol, and for O it's 16.00 g/mol. Thus, the molar mass of CaO is calculated as:
\[ 40.08\text{ g/mol} + 16.00 \text{ g/mol} = 56.08 \text{ g/mol} \]
This value is essential for converting between grams and moles in stoichiometric calculations.

Limiting Reactant

The limiting reactant is the substance that gets completely consumed first during a chemical reaction, thus determining the maximum amount of product that can be formed. To identify the limiting reactant, you must compare the molar amounts of reactants based on the balanced chemical equation. In the given reaction, 2 moles of calcium (Ca) react with 1 mole of oxygen (O₂) to produce 2 moles of calcium oxide (CaO). By calculating the moles of each reactant, you can see which one runs out first. For instance, if you have 0.1047 moles of Ca and 0.0875 moles of O₂, calcium is the limiting reactant because it produces fewer moles of CaO than oxygen.

Chemical Equations

Chemical equations represent the reactants and products in a chemical reaction. They must be balanced, meaning the number of atoms of each element is the same on both sides of the equation. This reflects the conservation of mass principle. In the reaction between calcium and oxygen, the balanced equation is:
\[ 2\text{Ca}(s) + \text{O}_2(g) \rightarrow 2\text{CaO}(s) \]
Here, 2 moles of calcium react with 1 mole of oxygen to form 2 moles of calcium oxide. Balancing the chemical equation ensures that stoichiometric calculations, like determining moles or masses of reactants and products, are accurate.

Mass Conversion

Mass conversion in stoichiometry involves converting between grams and moles using the molar mass. To find the number of moles from a given mass, use the formula:
\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
For example, to find the moles of calcium (Ca) from 4.20 g:
\[ \text{moles of Ca} = \frac{4.20 \text{ g}}{40.08 \text{ g/mol}} = 0.1047 \text{ mol} \]
Similarly, for oxygen (O₂) from 2.80 g:
\[ \text{moles of O}_2 = \frac{2.80 \text{ g}}{32.00 \text{ g/mol}} = 0.0875 \text{ mol} \]
These conversions are essential for performing subsequent stoichiometric calculations, such as identifying the limiting reactant and determining product yield.

Reaction Yields

Reaction yield is the amount of product formed in a chemical reaction, usually expressed in grams. To calculate the theoretical yield, use the moles of the limiting reactant and multiply by the molar mass of the product. From our example, calcium (Ca) is the limiting reactant producing 0.1047 moles of calcium oxide (CaO). To find the mass of CaO produced:
\[ \text{mass of CaO} = 0.1047 \text{ mol} \times 56.08 \text{ g/mol} = 5.87 \text{ g} \]
This calculation shows the theoretical maximum mass of CaO that can be produced, given the initial amounts of reactants.