Question

What mass of magnesium metal reacts with sulfuric acid to produce \(225 \mathrm{~mL}\) of hydrogen gas at STP? $$\mathrm{Mg}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(aq)+\mathrm{H}_{2}(g)$$

Short answer

0.244 g of magnesium reacts.

Step-by-step solution

Understand the Reaction

The given chemical reaction is between magnesium (Mg) and sulfuric acid (H₂SO₄) to produce magnesium sulfate (MgSO₄) and hydrogen gas (H₂). The equation is balanced with a 1:1 molar ratio for reactants and products.

Use Ideal Gas Law at STP

Standard Temperature and Pressure (STP) conditions are 0°C (273.15 K) and 1 atm. At STP, 1 mole of any gas occupies 22.4 liters (22400 mL). We need to find the moles of hydrogen gas produced with the given volume of 225 mL.

Calculate Moles of Hydrogen Gas

Using the formula for moles at STP, \( \text{Moles of } H_2 = \frac{\text{Volume of } H_2}{22400 \, \text{mL/mol}} \). Substitute the given volume: \( \text{Moles of } H_2 = \frac{225}{22400} \approx 0.01004 \text{ moles} \).

Determine Moles of Magnesium

The balanced equation shows a 1:1 molar ratio between magnesium and hydrogen gas, so \( 0.01004 \text{ moles of } H_2 \) requires \(0.01004 \text{ moles of } \text{Mg} \).

Calculate Mass of Magnesium

The molar mass of magnesium (Mg) is 24.31 g/mol. Calculate the mass using the moles found: \( \text{Mass of } Mg = 0.01004 \text{ moles} \times 24.31 \text{ g/mol} \approx 0.244 \text{ g} \).

Key concepts

Chemical Reaction

A chemical reaction typically involves the transformation of reactants into products, involving changes at the atomic level. In our exercise, magnesium metal reacts with sulfuric acid. This particular reaction can be represented by the equation: \[ \text{Mg}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{MgSO}_4(aq) + \text{H}_2(g) \]Here, magnesium and sulfuric acid are the reactants, and they form magnesium sulfate and hydrogen gas. This equation is balanced, meaning that the number of each type of atom on the reactants side matches the number on the products side. This balance is crucial because of the law of conservation of mass—matter cannot be created or destroyed in a chemical reaction.

Ideal Gas Law

The ideal gas law is a fundamental relationship between the pressure, volume, temperature, and quantity of a gas. It's represented by the equation: \( PV = nRT \), where:

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

In our problem, we consider Standard Temperature and Pressure (STP), where the temperature is 0°C or 273.15 K, and pressure is 1 atm. Under these conditions, one mole of an ideal gas occupies 22.4 liters. This principle helps us convert the volume of gas to moles, which is instrumental in finding the amount of other reacting substances.

Molar Mass

Molar mass is a measure that indicates the mass of one mole of a substance, typically expressed in grams per mole (g/mol). The molar mass of an element can be found on the periodic table, where it is usually indicated as the atomic mass. For magnesium, this value is 24.31 g/mol. Having the molar mass is crucial when converting between the mass of a substance and the quantity in moles. Since moles relate directly to particles involved in reactions, knowing the molar mass allows us to calculate how much of a substance we have or need in a reaction. In this exercise, the molar mass of magnesium helps us convert from moles to grams, allowing us to find the mass of magnesium metal needed.

Gas Volume Calculations

Calculating the volume of gases involves using their relationship with moles, especially under known conditions like STP (Standard Temperature and Pressure). At STP, the volume of one mole of any gas is 22.4 liters, equivalent to 22400 milliliters.To find the moles of hydrogen gas from a given volume, we apply the formula: \[ \text{Moles of gas} = \frac{\text{Volume of the gas}}{22400 \, \text{mL/mol}} \]In this problem, the provided volume of hydrogen is 225 mL, which when used in this formula, allows us to determine the moles of hydrogen gas produced. This linkage between volume and moles is central to many stoichiometric calculations involving gases.