91Ó°ÊÓ

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)â‚‚, and sulfuric acid, Hâ‚‚SOâ‚„. 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)â‚‚ = 0.0171
• Moles of Hâ‚‚SOâ‚„ = 0.00617

Since Hâ‚‚SOâ‚„ has fewer moles than Mg(OH)â‚‚, it is the limiting reactant. This means Hâ‚‚SOâ‚„ will run out first, thus limiting the amount of magnesium sulfate (MgSOâ‚„) produced.

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)â‚‚, Hâ‚‚SOâ‚„, and MgSOâ‚„.

• For Mg(OH)â‚‚: 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 Hâ‚‚SOâ‚„: Sum the atomic masses of hydrogen, sulfur, and oxygen to get 98.09 g/mol.
• For MgSOâ‚„: 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 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 (MgSOâ‚„) 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.