Impure phosphoric acid for use in the manufacture of fertilizers is produced by the reaction of sulfuric acid on phosphate rock, of which a principal component is \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)\). The reaction is $$ \begin{aligned} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s}) &+3 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \\ & 3 \mathrm{CaSO}_{4}(\mathrm{~s})+2 \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \end{aligned} $$ (a) How many moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) can be produced from the reaction of \(200 \mathrm{~kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ? (b) Determine the mass of calcium sulfate that is produced as a by. product of the reaction of \(200 \mathrm{~mol} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Short Answer

Expert verified
a) 1,359.50 moles of H3PO4; b) 81.684 Kg of CaSO4

Step by step solution

01

- Calculate the molar mass of H2SO4

The molar mass is calculated by summing the masses of all the atoms in a molecule of H2SO4. The atomic masses are approximately: H (1.01 g/mol), S (32.07 g/mol), and O (16.00 g/mol). Thus, the molar mass of H2SO4 is: 2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 2.02 g/mol + 32.07 g/mol + 64.00 g/mol = 98.09 g/mol.
02

- Convert the mass of H2SO4 to moles

Use the molar mass of H2SO4 from Step 1 to convert 200 kg of H2SO4 to moles. First, convert the mass from kilograms to grams: 200 kg * 1,000 g/kg = 200,000 g Then, use the molar mass to find the moles: 200,000 g / 98.09 g/mol = 2,039.25 moles of H2SO4.
03

- Apply stoichiometry to find moles of H3PO4

Use the balanced reaction to find the stoichiometric ratio between H2SO4 and H3PO4. From the equation, 3 moles of H2SO4 produce 2 moles of H3PO4. Set up a proportion:(3 moles H2SO4 / 2 moles H3PO4) = (2,039.25 moles H2SO4 / x moles H3PO4) Solve for x, which gives:x = (2 moles H3PO4 / 3 moles H2SO4) * 2,039.25 moles H2SO4 x = 1,359.50 moles H3PO4
04

- Calculate the mass of calcium sulfate (CaSO4)

For part b, given there are 200 moles of Ca3(PO4)2, we apply stoichiometry using the balanced equation. From the equation, 1 mole of Ca3(PO4)2 produces 3 moles of CaSO4. Therefore, 200 moles of Ca3(PO4)2 produces 600 moles of CaSO4. Calculate the molar mass of CaSO4: Ca (40.08 g/mol), S (32.07 g/mol) and O (16.00 g/mol). Thus, the molar mass of CaSO4 is:(40.08 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 136.14 g/molThen find the mass:600 moles * 136.14 g/mol = 81,684 g or 81.684 kg.

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Key Concepts

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

Molar Mass Calculation
Understanding molar mass is essential in the study of chemistry, as it allows for the conversion between grams and moles, two units commonly used to measure chemical substances. The molar mass of a substance is the weight of one mole (6.022 x 1023 particles) of that substance and is expressed in grams per mole (g/mol). To calculate the molar mass, one must sum the atomic masses of all the atoms present in a molecule.

For sulfuric acid (H2SO4), this involves using the atomic masses for hydrogen (H), sulfur (S), and oxygen (O). For each atom, multiply its atomic mass by the number of times that atom appears in the molecule and add the results together, as performed in the step by step solution. Knowing the molar mass allows chemists to relate mass to the amount of a substance in a practical way, which is crucial for all subsequent stoichiometry calculations.
Chemical Reaction Stoichiometry
Stoichiometry is the section of chemistry that pertains to the measurement of the quantitive relationships, or ratios, in which chemical substances react and form products. It directly uses the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

In the exercise given, stoichiometry is used to calculate how many moles of phosphoric acid (H3PO4) can be produced from a certain amount of sulfuric acid (H2SO4). This process requires understanding and applying the balanced chemical equation that represents the reaction. Each coefficient in the chemical equation gives the ratio of moles needed or produced of each compound. Incorrect stoichiometric calculations can lead to significant errors in chemical reactions, which is why it's critical to apply these principles accurately as demonstrated in the provided solution.
Mole-to-Mass Conversion
The mole-to-mass conversion is a fundamental concept in stoichiometry that enables chemists to calculate the mass of a substance when given the amount in moles. After using the molar mass for calculating how many moles of substance are present, you can convert that amount into grams through multiplication by the molar mass.

Once the molar mass is accurately determined, as seen in our exercise with sulfuric acid and calcium sulfate (CaSO4), it's simply a matter of applying the conversion factor to switch between moles and grams. The solution illustrated shows how to use the mole-to-mass conversion to determine the amount of byproduct, calcium sulfate, created during the production of phosphoric acid. This step is crucial for understanding the yields of reactions and for planning and optimizing chemical processes.
Phosphoric Acid Production
Phosphoric acid (H3PO4) is commonly produced through the reaction of sulfuric acid with phosphate rock. In the industrial synthesis of phosphoric acid, many factors including stoichiometry, purity of the reactants, and reaction conditions play a critical role.

The production process often generates byproducts, as shown in the chemical equation of the exercise, where calcium sulfate (CaSO4) is also formed. The stoichiometry explained in previous sections aids in predicting the amounts of products and byproducts, facilitating its application in real-world chemical industry scenarios. A thorough understanding of these principles is essential for efficient and practical chemical manufacturing, as it ensures that a chemical reaction will yield the right amount of product, which in this case is phosphoric acid, vital for the fertilizer industry.

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Most popular questions from this chapter

Potassium superoxide, \(\mathrm{KO}_{2}\), is utilized in closedsystem breathing apparatus to remove carbon dioxide and water from exhaled air. The removal of water generates oxygen for breathing by the reaction $$ 4 \mathrm{KO}_{2}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 3 \mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{KOH}(\mathrm{s}) $$ The potassium hydroxide removes carbon dioxide from the apparatus by the reaction $$ \mathrm{KOH}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{KHCO}_{3}(\mathrm{~s}) $$ (a) What mass of potassium superoxide generates \(20.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? (b) What mass of \(\mathrm{CO}_{2}\) can be removed from the apparatus by \(100 \mathrm{~g}\) of \(\mathrm{KO}_{2}\) ?

A 15.00-mL. sample of sodium hydroxide was titrated to the stoichiometric point with \(17.40 \mathrm{~mL}\). of \(0.234 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\). (a) What is the initial molarity of \(\mathrm{NaOH}\) in the solution? (b) Calculate the mass of \(\mathrm{NaOH}\) in the solution.

In a titration, a \(3.25 \mathrm{~g}\) sample of an acid, HX, requires \(68.8 \mathrm{~mL}\) of a \(0.750 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) solution for complete reaction. What is the molar mass of the acid?

A 15.00-mL. sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (with two acidic protons), was titrated to the stoichiometric point with \(17.02 \mathrm{~mL}\) of \(0.288 \mathrm{M} \mathrm{NaOH}(a q)\). (a) What is the molarity of the oxalic acid? (b) Dctermine the mass of oxalic acid in the solution.

\(\mathbf{L} .4\) The compound diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}\), was at one time considered for use as a rocket fuel. Its combustion reaction is $$ \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{l}) \rightarrow 2 \mathrm{HBO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$ The fact that \(\mathrm{HBO}_{2}\), a reactive compound, was produced rather than the relatively inert \(\mathrm{B}_{2} \mathrm{O}_{3}\) was a factor in the discontinuation of the investigation of diborane as a fuel. (a) What mass of liquid oxygen (LOX) would be needed to burn \(50.0 \mathrm{~g}\) of \(\mathrm{B}_{2} \mathrm{H}_{6}\) ? (b) Determine the mass of \(\mathrm{HBO}_{2}\) produced from the combustion of \(30.0 \mathrm{~g}\) of \(\mathrm{B}_{2} \mathrm{H}_{6}\).

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