
Balance the following equation.
\[CuS{O_4} + NaOH \to Cu{(OH)_2} + N{a_{2}}S{O_4}\]
Answer
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Hint: On the reactant and product sides of a balanced chemical equation, there are equivalent numbers of atoms for each element involved in the reaction. This is a condition that the equation must meet in order to be consistent with the conservation of matter law. It can be checked by adding the numbers of atoms to both sides of the arrow and comparing them to ensure that they are identical.
Complete answer:
You must ensure that the number of atoms of each element on the reactant side equals the number of atoms of each element on the product side in order to match the chemical equation. To make all sides equal, multiply the number of atoms in each product until they are equal on both sides.
The chemical formulae of the reactants and products must be used to create the unbalanced equation (if it is not already provided).
On the reactant and product sides, the total number of atoms of each element must be compared.
Stoichiometric coefficients are now applied to molecules that contain an element with a different number of atoms on the reactant and component sides.
The number of atoms on either side must be balanced by the coefficient.
The stoichiometric coefficients are typically given last to hydrogen and oxygen atoms.
The number of atoms in the reactant and component components must now be modified.
It's necessary to remember that the stoichiometric coefficient must be multiplied by the total number of atoms of the element contained in 1 molecule of the species to get the number of atoms in that species.
This Step is repeated until all of the reacting elements' atom counts on the reactant and product sides are equal.
\[CuS{O_4} + NaOH \to Cu{(OH)_2} + N{a_{2}}S{O_4}\]
NaOH – Multiply by 2 stoichiometrically we get,
\[CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_{2}}S{O_4}\]
Note:
Note that the number of atoms in any formula containing the element is determined by multiplying the coefficient by the element's subscript in the formula. If an element occurs in several formulas on one side of an equation, the number of atoms represented in each must be calculated and then added.
Complete answer:
You must ensure that the number of atoms of each element on the reactant side equals the number of atoms of each element on the product side in order to match the chemical equation. To make all sides equal, multiply the number of atoms in each product until they are equal on both sides.
The chemical formulae of the reactants and products must be used to create the unbalanced equation (if it is not already provided).
On the reactant and product sides, the total number of atoms of each element must be compared.
Stoichiometric coefficients are now applied to molecules that contain an element with a different number of atoms on the reactant and component sides.
The number of atoms on either side must be balanced by the coefficient.
The stoichiometric coefficients are typically given last to hydrogen and oxygen atoms.
The number of atoms in the reactant and component components must now be modified.
It's necessary to remember that the stoichiometric coefficient must be multiplied by the total number of atoms of the element contained in 1 molecule of the species to get the number of atoms in that species.
This Step is repeated until all of the reacting elements' atom counts on the reactant and product sides are equal.
\[CuS{O_4} + NaOH \to Cu{(OH)_2} + N{a_{2}}S{O_4}\]
| Reactant side | Product side | |
| Cu | 1 | 1 |
| SO₄ | 1 | 1 |
| Na | 1 | 2 |
| OH | 1 | 2 |
NaOH – Multiply by 2 stoichiometrically we get,
| Reactant side | Product side | |
| Cu | 1 | 1 |
| $SO_4$ | 1 | 1 |
| Na | 2 | 2 |
| OH | 2 | 2 |
\[CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_{2}}S{O_4}\]
Note:
Note that the number of atoms in any formula containing the element is determined by multiplying the coefficient by the element's subscript in the formula. If an element occurs in several formulas on one side of an equation, the number of atoms represented in each must be calculated and then added.
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