Question

Clay contains \(30 \% \mathrm{AI}_{2} \mathrm{O}_{3}, 55 \% \mathrm{SiO}_{2}\), and \(15 \% \mathrm{H}_{2} \mathrm{O}\). What weight of limestone is required per ton of clay to carry out the following chemical change? \(6 \mathrm{CaCO}_{3}+\mathrm{AI}_{2} \mathrm{O}_{3}+\mathrm{SiO}_{2} \rightarrow 3 \mathrm{CaO} \cdot \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{CaO}-\mathrm{SiO}_{2}+6 \mathrm{CO}_{2}\) (limestone)

Short answer

The weight of limestone required per ton of clay is approximately 7,259 kg.

Step-by-step solution

Find the moles of Al2O3 and SiO2 in one ton of clay

Since clay contains 30% Al2O3 and 55% SiO2, we need to find the moles of Al2O3 and SiO2 in one ton of clay. Since one ton is equal to 1000 kg, we have: - Weight of Al2O3 = 0.30 * 1000 kg = 300 kg - Weight of SiO2 = 0.55 * 1000 kg = 550 kg Now, we find the moles of each compound using their respective molar masses: - Moles of Al2O3 = 300 kg / (2 * 26.98 + 3 * 16.00) g/mol = 300000 g / 101.96 g/mol = 2941.4 mol - Moles of SiO2 = 550 kg / (28.09 + 2 * 16.00) g/mol = 550000 g / 60.09 g/mol = 9151.6 mol

Determine the amount of limestone needed according to the mole ratio in the balanced chemical equation

In the given balanced chemical equation, we see that 6 moles of CaCO3 (limestone) react with 1 mole of Al2O3 and 1 mole of SiO2. Based on the moles of each calculated in the previous step, we determine the moles of limestone needed: Moles of limestone (CaCO3) needed = 6 * (moles of Al2O3 + moles of SiO2) = 6 * (2941.4 + 9151.6) = 72558 mol

Convert moles of limestone to weight

Now that we have the moles of limestone needed, we can determine the weight of limestone required for one ton of clay using the molar mass of CaCO3: Weight of limestone (CaCO3) = moles of limestone * molar mass of CaCO3 = 72558 mol * (40.08 + 12.01 + 3*16.00) g/mol = 72558 * 100.09 g/mol = 7258792 g

Report the final result

The weight of limestone required per ton of clay is 7,258,792 grams or approximately 7,259 kg.

Key concepts

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is essential for solving problems like the one presented. Simply put, stoichiometry is the part of chemistry that deals with the relationships between the quantities of reactants and products in a chemical reaction. It is grounded in the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

To work through stoichiometry problems, the first step is always to ensure you have a balanced chemical equation. This tells you the mole ratio of each reactant and product involved. For instance, the given chemical equation demonstrates that six moles of limestone react with one mole of aluminum oxide and one mole of silicon dioxide.

Mole Ratio

In this example, we see a direct application of stoichiometry. We use the mole ratio derived from the balanced equation to calculate the quantity of limestone needed. After finding the moles of aluminum oxide and silicon dioxide in a ton of clay, we use their sum multiplied by the mole ratio from the equation (6 moles of limestone to 1 mole of aluminum oxide and silicon dioxide) to find the required moles of limestone.

Ultimately, stoichiometry serves as the bridge between the molecular world and the real-world, translating moles into tangible quantities like grams that can be measured and used in practical scenarios.

Molar Mass Calculation

The concept of molar mass is a fundamental aspect of stoichiometry and is crucial for converting grams to moles and vice versa. The molar mass is the weight of one mole of a substance and it is expressed in grams per mole (g/mol). Mathematically, the molar mass of a compound is calculated by summing the atomic masses of each element, as found on the periodic table, multiplied by the number of atoms of that element in the formula.

Finding Molar Mass

In the exercise, we calculated the molar mass for aluminum oxide (Al_{2}O_{3}) as well as silicon dioxide (SiO_{2}), which are integral to determining the number of moles of these compounds in one ton of clay. The molar mass is used to convert the weight of a compound from grams to moles. This step is significant since stoichiometry calculations are performed in moles. The balance and precision of molar mass calculations ensure accuracy when translating from the molecular scale to practical quantities, such as in the amount of limestone needed for the reaction.

Balanced Chemical Equations

Balancing chemical equations is a necessary skill in chemistry that ensures the principle of the conservation of mass is satisfied. A balanced equation has equal numbers of each type of atom on both sides of the reaction. This allows chemists to see, at a glance, the mole ratios of reactants and products, forming the basis for all stoichiometric calculations.

Importance of Equation Balance

When we look at the balanced chemical equation in the exercise, it shows a complex reaction involving limestone, aluminum oxide, and silicon dioxide. Getting to a balanced equation involves ensuring that the number of atoms for each element is the same before and after the reaction. Without a balanced equation, it is impossible to correctly apply stoichiometric principles to calculate reactant or product quantities.By meticulously verifying the atomic balance, chemists can approach problems methodically, ensuring that the stoichiometric coefficients reflect the true proportions in which substances react. This makes it possible to predict quantities needed or produced, as illustrated by the accurate calculation of limestone required in the chemical change involving clay.