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Consider the following unbalanced equation: $$ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{~kg}\) calcium phosphate with \(1.0 \mathrm{~kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}, \mathrm{SO}_{4}\right.\) by mass)?

Short Answer

Expert verified
In this balanced chemical equation, $$2 \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) + 6 \mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow 6 \mathrm{CaSO}_{4}(s) + 4 \mathrm{H}_{3}\mathrm{PO}_{4}(aq)$$, from 1.0 kg of calcium phosphate and 1.0 kg of concentrated sulfuric acid (98% by mass), 1315.87 g of calcium sulfate (CaSO鈧) and 631.12 g of phosphoric acid (H鈧働O鈧) can be produced.

Step by step solution

01

Balance the chemical equation

Balancing the given chemical equation, we get: $$ 2 \mathrm{Ca}_{3}(\mathrm{PO}_{4})_{2}(s) + 6 \mathrm{H}_{2}\mathrm{SO}_{4}(aq) \longrightarrow 6 \mathrm{CaSO}_{4}(s) + 4 \mathrm{H}_{3}\mathrm{PO}_{4}(aq) $$
02

Convert the given masses into moles

We need the molar masses of the substances involved: - Ca鈧(PO鈧)鈧: Molar mass = 3(40.08) + 2(1.0079)(4) + 2(30.97)(2) 鈮 310.18 g/mol - H鈧係O鈧: Molar mass = 2(1.0079) + 32.07 + 4(16.00) 鈮 98.08 g/mol Now we can convert the given masses of 1.0 kg each into moles: - For Ca鈧(PO鈧)鈧: \( \frac{1000 \text{ g}}{310.18 \text{ g/mol}} 鈮 3.22 \text{ mol} \) of Ca鈧(PO鈧)鈧 - For H鈧係O鈧: since it's 98% by mass, the mass of H鈧係O鈧 in 1.0 kg of concentrated solution is 0.98 脳 1000 g = 980 g. So, \( \frac{980 \text{ g}}{98.08 \text{ g/mol}} 鈮 9.99 \text{ mol} \) of H鈧係O鈧
03

Determine the limiting reagent

Compare the mole ratio of the reacted substances to the balanced chemical equation: Ca鈧(PO鈧)鈧: \(\frac{3.22 \text{ mol}}{2} = 1.61 \) H鈧係O鈧: \(\frac{9.99 \text{ mol}}{6} = 1.67 \) Since the Ca鈧(PO鈧)鈧 mole ratio is smaller, Ca鈧(PO鈧)鈧 is the limiting reagent.
04

Calculate the number of moles of the products formed

From the balanced chemical equation, we can see the mole ratio of the reactants with respect to the products: 1. Ca鈧(PO鈧)鈧:CaSO鈧 = 2:6 2. Ca鈧(PO鈧)鈧:H鈧働O鈧 = 2:4 Given 3.22 mol Ca鈧(PO鈧)鈧: - Moles of CaSO鈧 produced = 6 脳 (3.22 mol Ca鈧(PO鈧)鈧 / 2) = 9.66 mol CaSO鈧 - Moles of H鈧働O鈧 produced = 4 脳 (3.22 mol Ca鈧(PO鈧)鈧 / 2) = 6.44 mol H鈧働O鈧
05

Convert the moles of the products back into masses

We need the molar masses of the products: - CaSO鈧: Molar mass = 40.08 + 32.07 + 4(16.00) 鈮 136.14 g/mol - H鈧働O鈧: Molar mass = 3(1.0079) + 31.00 + 4(16.00) 鈮 98.00 g/mol Now we can convert moles of CaSO鈧 and H鈧働O鈧 back into their respective masses: - Mass of CaSO鈧 produced = 9.66 mol 脳 136.14 g/mol 鈮 1315.87 g - Mass of H鈧働O鈧 produced = 6.44 mol 脳 98.00 g/mol 鈮 631.12 g Therefore, 1315.87 g of calcium sulfate (CaSO鈧) and 631.12 g of phosphoric acid (H鈧働O鈧) can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantifiable relationships between the reactants and products in a chemical reaction. It is rooted in the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction. Consequently, the amount of each element must be the same in the products as it is in the reactants.

Understanding stoichiometry involves a series of steps. It begins with a balanced chemical equation, which ensures that the number of atoms for each element is the same on both sides of the reaction. Once the equation is balanced, you can use it as a map from which to determine the proportion of reactants and products involved.

In the provided exercise, stoichiometry helps us to utilize the balanced chemical equation to find out the exact amounts of calcium sulfate and phosphoric acid produced when 1.0 kg of calcium phosphate reacts with 1.0 kg of sulfuric acid.
Limiting Reagent
The concept of the limiting reagent is central to quantitative chemistry and stoichiometry. In essence, the limiting reagent is the substance that is used up first in a reaction, determining the maximum amount of product that can be formed. When it is exhausted, the reaction stops, regardless of how much of the other reactants are left.

Identifying the limiting reagent requires comparing the mole ratio of the available reactants to their ratios in the balanced chemical equation. In the textbook exercise, the chemical equation indicated that two moles of calcium phosphate could react with six moles of sulfuric acid to produce products. By calculating the amounts of each reagent available in moles, the equation showed that calcium phosphate was the limiting reagent because it had the smaller mole ratio. This analysis informs us that the amount of product will be based on the initial amount of calcium phosphate present.
Molar Mass Calculations
Molar mass calculations are critical for converting between the mass of a substance and the amount in moles, as moles are the units used in chemical equations. The molar mass is the weight of one mole of a substance and is measured in grams per mole (g/mol). It can be calculated by summing the atomic masses of all the atoms in a molecule as listed on the periodic table.

In the solution to the exercise, molar masses were necessary to convert the given masses (in kilograms) of calcium phosphate and sulfuric acid into moles. These mole quantities were then used to determine the mass of the products formed in the reaction. Without accurate molar mass calculations, it would not be possible to accurately predict the masses of products formed.

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Most popular questions from this chapter

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