Question

If \(1.00 \mathrm{~g}\) of magnesium hydroxide reacts with \(0.605 \mathrm{~g}\) of sulfuric acid, what is the mass of magnesium sulfate produced? $$\mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{MgSO}_{4(s)+\mathrm{H}_{2} \mathrm{O}(l)$$

Short answer

The mass of magnesium sulfate produced is 0.743 grams.

Step-by-step solution

Determine molar masses

Calculate the molar mass of magnesium hydroxide (Mg(OH)₂), sulfuric acid (H₂SO₄), and magnesium sulfate (MgSO₄). - Molar mass of Mg(OH)₂ = 24.31 (Mg) + 2(16.00 + 1.01) = 58.32 g/mol - Molar mass of H₂SO₄ = 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol - Molar mass of MgSO₄ = 24.31 + 32.07 + 4(16.00) = 120.38 g/mol.

Calculate moles of reactants

Determine the number of moles of each reactant. - Moles of Mg(OH)₂ = \( \frac{1.00 \text{ g}}{58.32 \text{ g/mol}} = 0.0171 \text{ moles} \)- Moles of H₂SO₄ = \( \frac{0.605 \text{ g}}{98.09 \text{ g/mol}} = 0.00617\text{ moles} \)

Identify the limiting reactant

Since the reaction equation is balanced with a 1:1 mole ratio, compare the moles of Mg(OH)₂ and H₂SO₄. - Mg(OH)₂: 0.0171 moles - H₂SO₄: 0.00617 moles Thus, H₂SO₄ is the limiting reactant as it has fewer moles available to react.

Calculate moles of magnesium sulfate produced

Use the moles of the limiting reactant (H₂SO₄) to find the moles of magnesium sulfate produced. - Since the reaction produces 1 mole of MgSO₄ for every 1 mole of H₂SO₄, the moles of MgSO₄ will also be 0.00617.

Calculate mass of magnesium sulfate produced

Finally, convert moles of magnesium sulfate to grams using its molar mass.- Mass of MgSO₄ = \( 0.00617 \text{ moles} \times 120.38 \text{ g/mol} = 0.743 \text{ grams} \).

Key concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that determines the amount of product that can be formed. It is the reactant that gets completely used up first, ensuring the reaction cannot continue further. In this particular exercise, we consider magnesium hydroxide, Mg(OH)₂, and sulfuric acid, H₂SO₄. The balanced equation shows a 1:1 ratio between these reactants.

To identify the limiting reactant, we first calculate the number of moles of each reactant separately. Using their respective molar masses, as given in our exercise solution, we calculate:

• Moles of Mg(OH)₂ = 0.0171
• Moles of H₂SO₄ = 0.00617

Since H₂SO₄ has fewer moles than Mg(OH)₂, it is the limiting reactant. This means H₂SO₄ will run out first, thus limiting the amount of magnesium sulfate (MgSO₄) produced.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of that substance. To calculate the molar mass, we add together the atomic masses of all atoms present in a molecule. This is essential for converting between grams and moles in stoichiometry. In our solved problem, the molar mass calculations are applied for three substances: Mg(OH)₂, H₂SO₄, and MgSO₄.

• For Mg(OH)₂: Add the atomic mass of magnesium (24.31 g/mol), oxygen (16.00 g/mol), and hydrogen (1.01 g/mol), calculated as 58.32 g/mol.
• For H₂SO₄: Sum the atomic masses of hydrogen, sulfur, and oxygen to get 98.09 g/mol.
• For MgSO₄: Mix the atomic masses of magnesium, sulfur, and oxygen amounting to 120.38 g/mol.

These calculations allow us to proceed with converting given grams of reactants into moles, which is crucial to determine the limiting reactant and predict the quantity of product formed.

Chemical Equations

Chemical equations are symbolic representations showing the substances involved in a chemical reaction and their stoichiometric relationships. They are balanced to reflect the conservation of mass, where the same number of each type of atom must appear on both sides of the equation.

In the reaction equation from the exercise: \[\mathrm{Mg(OH)}_{2}(s)+\mathrm{H}_{2}\mathrm{SO}_{4}(l) \rightarrow \mathrm{MgSO}_{4(s)+\mathrm{H}_{2}\mathrm{O}(l}\]Each side of this equation must have the same number of atoms for every element involved:

• Magnesium (Mg), Sulfur (S), Oxygen (O), and Hydrogen (H) are all balanced, with a 1:1:4:4 ratio represented on both sides.
• Stoichiometry allows us to translate this balanced equation into predictions of how much product (MgSO₄) will be formed when specific amounts of reactants are used.

Understanding stoichiometric coefficients is key; in this example, a 1:1:1 ratio demonstrates that two moles of reactants yield two moles of products, applied directly to compute product masses.