Magnesium Oxide Mass And Oxygen Volume Calculation In Mg And O₂ Reaction

JU07/16/2025 08, 2025 by THE IDEN 73 views

Calculating Magnesium Oxide Mass and Oxygen Volume in the Reaction of Magnesium and Oxygen

In this comprehensive guide, we will delve into the fascinating world of stoichiometry and explore the reaction between magnesium and oxygen. Stoichiometry, the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical reaction, is a fundamental concept in chemistry. Our primary objective is to calculate the mass of magnesium oxide (MgO) produced from 16 g of magnesium (Mg) in the reaction between magnesium and oxygen (O₂). Additionally, we will determine the volume of oxygen at Standard Temperature and Pressure (STP) that reacts with the same amount of magnesium. This exploration will involve balancing the chemical equation, applying molar mass concepts, and utilizing the ideal gas law. Understanding these principles is crucial for comprehending chemical reactions and their quantitative aspects.

The first crucial step in any stoichiometry problem is to ensure that the chemical equation is correctly balanced. The unbalanced reaction provided is:

Mg(s) + O₂(g) → MgO(s)

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. By inspection, we can see that there are two oxygen atoms on the reactant side (O₂) and only one oxygen atom on the product side (MgO). To balance the oxygen atoms, we place a coefficient of 2 in front of MgO:

Mg(s) + O₂(g) → 2 MgO(s)

Now, we have two oxygen atoms on each side. However, we now have two magnesium atoms on the product side and only one on the reactant side. To balance the magnesium atoms, we place a coefficient of 2 in front of Mg:

2 Mg(s) + O₂(g) → 2 MgO(s)

The equation is now balanced. We have two magnesium atoms and two oxygen atoms on both the reactant and product sides. The balanced chemical equation is essential because it provides the mole ratios between the reactants and products, which are crucial for stoichiometric calculations. This balanced equation tells us that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide. This ratio is the foundation for our subsequent calculations.

Next, we need to calculate the number of moles of magnesium (Mg) present in 16 g of magnesium. To do this, we use the molar mass of magnesium. The molar mass of an element is the mass of one mole of that element, usually expressed in grams per mole (g/mol). You can find the molar mass of magnesium on the periodic table, which is approximately 24.31 g/mol. This means that one mole of magnesium has a mass of 24.31 grams.

The formula to calculate the number of moles is:

Moles = Mass / Molar Mass

Given that we have 16 g of magnesium, we can plug the values into the formula:

Moles of Mg = 16 g / 24.31 g/mol

Moles of Mg ≈ 0.658 moles

Therefore, we have approximately 0.658 moles of magnesium. This value is crucial because it allows us to use the stoichiometric ratios from the balanced equation to determine the moles of other reactants and products involved in the reaction. Understanding the mole concept is fundamental in chemistry as it links the macroscopic property of mass to the microscopic world of atoms and molecules.

Now that we know the moles of magnesium, we can use the balanced chemical equation to determine the moles of magnesium oxide (MgO) produced. From the balanced equation:

2 Mg(s) + O₂(g) → 2 MgO(s)

We see that 2 moles of Mg react to produce 2 moles of MgO. This gives us a mole ratio of 1:1 between Mg and MgO. This ratio is the key to calculating how much MgO will be formed from the given amount of Mg. The stoichiometric coefficients in the balanced equation directly provide these mole ratios, which are essential for quantitative analysis in chemical reactions.

Using this ratio, we can calculate the moles of MgO produced:

Moles of MgO = Moles of Mg × (2 moles MgO / 2 moles Mg)

Moles of MgO = 0.658 moles × 1

Moles of MgO ≈ 0.658 moles

So, 0.658 moles of magnesium will produce approximately 0.658 moles of magnesium oxide. This direct relationship, derived from the balanced equation, simplifies the calculation and highlights the importance of stoichiometry in predicting product yields.

To find the mass of magnesium oxide (MgO) produced, we need to use the molar mass of MgO. The molar mass of MgO is the sum of the molar masses of magnesium (Mg) and oxygen (O). The molar mass of Mg is approximately 24.31 g/mol, and the molar mass of O is approximately 16.00 g/mol.

Molar mass of MgO = Molar mass of Mg + Molar mass of O

Molar mass of MgO = 24.31 g/mol + 16.00 g/mol

Molar mass of MgO ≈ 40.31 g/mol

Now that we have the molar mass of MgO and the moles of MgO produced (0.658 moles), we can calculate the mass of MgO using the formula:

Mass = Moles × Molar Mass

Mass of MgO = 0.658 moles × 40.31 g/mol

Mass of MgO ≈ 26.52 g

Therefore, approximately 26.52 grams of magnesium oxide is produced from 16 g of magnesium. This calculation demonstrates how molar mass and mole ratios are used in tandem to convert moles of a substance into mass, a crucial skill in quantitative chemistry.

To calculate the volume of oxygen required, we first need to determine the number of moles of oxygen (O₂) that react with 16 g of magnesium. From the balanced equation:

2 Mg(s) + O₂(g) → 2 MgO(s)

We see that 2 moles of Mg react with 1 mole of O₂. This gives us a mole ratio of 2:1 between Mg and O₂. This ratio is critical for determining how much oxygen is needed to react completely with the magnesium.

Using this ratio, we can calculate the moles of O₂ required:

Moles of O₂ = Moles of Mg × (1 mole O₂ / 2 moles Mg)

Moles of O₂ = 0.658 moles × (1/2)

Moles of O₂ ≈ 0.329 moles

Therefore, approximately 0.329 moles of oxygen are required to react with 16 g of magnesium. This calculation underscores the importance of using the correct stoichiometric ratios from the balanced equation to accurately determine reactant requirements.

To calculate the volume of oxygen at Standard Temperature and Pressure (STP), we use the ideal gas law. STP is defined as 0°C (273.15 K) and 1 atmosphere (atm) pressure. The ideal gas law is given by the equation:

PV = nRT

Where:

P is the pressure (in atm)
V is the volume (in liters)
n is the number of moles
R is the ideal gas constant (0.0821 L atm / (mol K))
T is the temperature (in Kelvin)

At STP, P = 1 atm and T = 273.15 K. We have already calculated the moles of O₂ (n ≈ 0.329 moles). Now, we can rearrange the ideal gas law equation to solve for V:

V = nRT / P

Plugging in the values:

V = (0.329 moles) × (0.0821 L atm / (mol K)) × (273.15 K) / (1 atm)

V ≈ 7.37 L

Therefore, approximately 7.37 liters of oxygen at STP are required to react with 16 g of magnesium. This calculation demonstrates the practical application of the ideal gas law in determining gas volumes under specific conditions.

In this comprehensive exploration, we successfully calculated the mass of magnesium oxide produced and the volume of oxygen required in the reaction between magnesium and oxygen. We meticulously balanced the chemical equation, determined the moles of reactants and products, and applied the ideal gas law to find the volume of oxygen at STP. Specifically, we found that approximately 26.52 grams of magnesium oxide are produced from 16 g of magnesium, and approximately 7.37 liters of oxygen at STP are required for the reaction. These calculations highlight the importance of stoichiometry and the ideal gas law in quantitative chemistry. By mastering these concepts, one can accurately predict the outcomes of chemical reactions and their quantitative aspects, which is fundamental to various scientific and industrial applications.

Keywords: Stoichiometry, magnesium oxide, magnesium, oxygen, balanced chemical equation, molar mass, ideal gas law, STP, moles, volume, mass, chemical reaction, quantitative chemistry.