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Balance tire chemical equation for each reaction. $$\mathrm{HClO}_{4}+\mathrm{P}_{4} \mathrm{O}_{10} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{Cl}_{2} \mathrm{O}_{7}$$

Short Answer

Expert verified
6\( \mathrm{HClO}_{4} \) + \( \mathrm{P}_{4} \mathrm{O}_{10} \) \( \rightarrow \) 4\( \mathrm{H}_{3} \mathrm{PO}_{4} \) + 3\( \mathrm{Cl}_{2} \mathrm{O}_{7} \).

Step by step solution

01

List Reactants and Products

Identify the elements present in the chemical equation:Reactants: Perchloric acid \( \mathrm{HClO}_{4} \) and phosphorus pentoxide \( \mathrm{P}_{4} \mathrm{O}_{10} \).Products: Phosphoric acid \( \mathrm{H}_{3} \mathrm{PO}_{4} \) and dichlorine heptoxide \( \mathrm{Cl}_{2} \mathrm{O}_{7} \).
02

Count Atoms for Each Element

In the initial unbalanced equation, count the number of atoms for each element present on both sides:- Left side: \( \mathrm{HClO}_{4} \) contains 1 H, 1 Cl, 4 O; \( \mathrm{P}_{4} \mathrm{O}_{10} \) contains 4 P, 10 O.- Right side: \( \mathrm{H}_{3} \mathrm{PO}_{4} \) contains 3 H, 1 P, 4 O; \( \mathrm{Cl}_{2} \mathrm{O}_{7} \) contains 2 Cl, 7 O.
03

Balance Phosphorus Atoms

To balance Phosphorus (P) atoms, ensure equal numbers on both sides. Currently, the left side has 4 P atoms while the right side has 1 P atom in \( \mathrm{H}_{3} \mathrm{PO}_{4} \). Add a coefficient of 4 to \( \mathrm{H}_{3} \mathrm{PO}_{4} \), resulting in:\[ \mathrm{P}_{4} \mathrm{O}_{10} + \mathrm{HClO}_{4} \rightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4} + \mathrm{Cl}_{2} \mathrm{O}_{7} \]
04

Balance Chlorine Atoms

To balance Chlorine (Cl) atoms, adjust to ensure equal numbers on both sides. The right side has 2 Cl atoms in \( \mathrm{Cl}_{2} \mathrm{O}_{7} \), so place a coefficient of 2 before \( \mathrm{HClO}_{4} \), yielding:\[ 2 \mathrm{HClO}_{4} + \mathrm{P}_{4} \mathrm{O}_{10} \rightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4} + \mathrm{Cl}_{2} \mathrm{O}_{7} \]
05

Balance Oxygen Atoms

Balance the oxygen atoms. Currently, there are a total of 18 O atoms on both sides (Left: 8 from 2 \( \mathrm{HClO}_{4} \) and 10 from \( \mathrm{P}_{4} \mathrm{O}_{10} \); Right: 16 from 4 \( \mathrm{H}_{3} \mathrm{PO}_{4} \) and 14 from \( \mathrm{Cl}_{2} \mathrm{O}_{7} \)). However, we need to consider the hydrogens next to confirm.
06

Balance Hydrogen Atoms

Check the hydrogen balance. The left side has 2 H atoms (from \( 2 \mathrm{HClO}_{4} \)) and the right side has 12 H atoms from \( \mathrm{H}_{3} \mathrm{PO}_{4} \). Add a coefficient of 6 to \( \mathrm{HClO}_{4} \) to balance the hydrogen atoms:\[ 6 \mathrm{HClO}_{4} + \mathrm{P}_{4} \mathrm{O}_{10} \rightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4} + 3 \mathrm{Cl}_{2} \mathrm{O}_{7} \]
07

Verify the Balance of All Atoms

After adjusting, check if all atoms are balanced: - P: 4 atoms on both sides. - Cl: 6 atoms on both sides. - O: 28 atoms on both sides. - H: 12 atoms on both sides. The equation is now balanced.

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Key Concepts

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

Chemical Reactions
Chemical reactions are fascinating processes where substances, known as reactants, are transformed into new substances, called products. This transformation involves the breaking and formation of bonds between atoms. Reactants and products are key terms that help us understand what is changing in a reaction. In our example, we have reactants such as perchloric acid (\( \mathrm{HClO}_4 \)) and phosphorus pentoxide (\( \mathrm{P}_4 \mathrm{O}_{10} \)), transforming into products like phosphoric acid (\( \mathrm{H}_3 \mathrm{PO}_4 \)) and dichlorine heptoxide (\( \mathrm{Cl}_2 \mathrm{O}_7 \)).
Studying chemical reactions require observing the rearrangement of atoms to ensure they are conserved, as dictated by the law of conservation of mass. This means that the atoms present in the reactants must be accounted for among the products.
Stoichiometry
Stoichiometry is the heart of balancing equations, providing the mathematical toolkit for achieving balance. This involves using coefficients to ensure each element's atom count is the same on both sides of the equation. It starts by understanding the molar ratios between reactants and products.
In our case, stoichiometry guided us to adjust the coefficients for each compound, such as adding a coefficient of 4 to \( \mathrm{H}_3 \mathrm{PO}_4 \) to match the 4 phosphorus atoms from \( \mathrm{P}_4 \mathrm{O}_{10} \) on the reactant side.
Understanding stoichiometry helps ensure that no atom goes missing during the reaction, maintaining equilibrium with respect to both the number and types of atoms involved. This careful balancing act ensures that the transformed substances retain their core atomic composition, just in a different structured form.
Atom Conservation
Atom conservation, as a principle, underscores the importance of maintaining an equal number of each type of atom on both sides of a chemical equation. This principle stems from the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a closed system.
In balancing our chemical equation, we counted and ensured there are equal numbers of hydrogen, chlorine, phosphorus, and oxygen atoms on both sides. For example, ensuring that 4 phosphorus atoms from \( \mathrm{P}_4 \mathrm{O}_{10} \) on the reactant side matched the 4 phosphorus atoms in the 4 \( \mathrm{H}_3 \mathrm{PO}_4 \) on the product side.
Atom conservation reminds us that while chemical reactions change the forms and combinations of substances, the fundamental amount of each kind of atom remains unchanged. This is crucial for accurately understanding chemical changes and ensuring reactions faithfully follow physical laws.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions where reactant and product species are depicted on the left and right sides separated by an arrow indicating the direction of the reaction. A properly balanced chemical equation mirrors the stoichiometry of the reaction.
In the given exercise, the equation started unbalanced, with unequal numbers of atoms on each side. By following a systematic balancing approach—adjusting coefficients—, we aligned the number of atoms for each element.
Writing a well-balanced chemical equation is essential for accurately representing the chemical changes occurring, ensuring that quantities can be experimentally verified, and facilitating precise communication in chemical sciences. In our case, transforming an unbalanced equation into a balanced form involved using coefficients to reflect accurate proportions, making sure the equation conforms to both stoichiometry and conservation laws.

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