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Predict the products of each precipitation reaction. Balance the equation, and then write the net ionic equation. (a) \(\mathrm{NiCl}_{2}(\mathrm{aq})+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(\mathrm{aq}) \rightarrow\) (b) \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \rightarrow\)

Short Answer

Expert verified
(a) Precipitate: NiS, Net ionic: \( \mathrm{Ni}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{NiS}\) (b) Precipitate: \(\mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}\), Net ionic: \(3\mathrm{Mn}^{2+} + 2\mathrm{PO}_{4}^{3-} \rightarrow \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}\)

Step by step solution

01

Predict Products for Reaction (a)

When mixing \( \mathrm{NiCl}_{2}(\mathrm{aq}) \) with \( \left(\mathrm{NH}_{4}\right)_{2}\mathrm{S}(\mathrm{aq}) \), the products are formed by exchanging the anions and cations. The products are \( \mathrm{NiS}(s) \) and \( \mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq}) \). \( \mathrm{NiS} \) is insoluble and precipitates.
02

Balance Equation for Reaction (a)

Balance the reaction: \[ \mathrm{NiCl}_{2}(\mathrm{aq}) + \left(\mathrm{NH}_{4}\right)_{2}\mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{NiS}(s) + 2\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq}) \] The equation is now balanced with the same number of each type of atom on both sides.
03

Net Ionic Equation for Reaction (a)

For the net ionic equation, remove spectator ions: \[ \mathrm{Ni}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{NiS}(s) \] This shows only the ions that form the precipitate.
04

Predict Products for Reaction (b)

In the reaction between \( \mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \) and \( \mathrm{Na}_{3}\mathrm{PO}_{4}(\mathrm{aq}) \), the products formed are \( \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}(s) \) and \( \mathrm{NaNO}_{3}(\mathrm{aq}) \). \( \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2} \) is a precipitate.
05

Balance Equation for Reaction (b)

Balance the reaction: \[ 3\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) + 2\mathrm{Na}_{3}\mathrm{PO}_{4}(\mathrm{aq}) \rightarrow \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}(s) + 6\mathrm{NaNO}_{3}(\mathrm{aq}) \] Now all atoms balance on both sides of the equation.
06

Net Ionic Equation for Reaction (b)

Remove spectator ions to form the net ionic equation: \[ 3\mathrm{Mn}^{2+}(\mathrm{aq}) + 2\mathrm{PO}_{4}^{3-}(\mathrm{aq}) \rightarrow \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}(s) \] This equation includes only the ions that directly form the precipitate.

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Key Concepts

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

Net Ionic Equation
When dealing with precipitation reactions, creating a net ionic equation is a powerful tool for understanding what's really happening at a molecular level. In a chemical reaction taking place in aqueous solution, ions are often shuffled through the disease of double replacement. However, not all of them result in a product that we are interested in. By focusing on the net ionic equation, you are essentially peeling away all the unimportant layers to see the core interaction that results in a new chemical compound.

For example, in reactions like the ones given in the exercise, the net ionic equation allows you to display only the ions that actually participate in the formation of the precipitate (the solid product that forms and settles out of the solution). Here's what you do:
  • Start by writing a balanced molecular equation.
  • Separate the aqueous compounds into their respective ions. Remember, only compounds that are aqueous will dissociate into ions.
  • Identify the spectator ions 鈥 these are ions that don't change during the reaction and can be canceled out from both sides of the equation.
  • What remains are the ions that directly combine to create the precipitate.
Through this process, you simplify the entire reaction to its essence, making it much easier to understand and analyze.
Solubility Rules
Understanding solubility rules is crucial for predicting whether a precipitation reaction will occur. Solubility rules are essentially guidelines that predict the solubility of different ionic compounds in water. These rules help determine which combinations of ions will form a precipitate, a solid, when mixed in an aqueous solution.

Here are the most commonly used solubility rules to remember:
  • Most nitrate (NO鈧冣伝), acetate (C鈧侶鈧僌鈧傗伝), and alkali metal (Li鈦, Na鈦, K鈦, etc.) salts are soluble.
  • Most chloride (Cl鈦), bromide (Br鈦), and iodide (I鈦) salts are soluble, except for those combined with Ag鈦, Pb虏鈦, and Hg鈧偮测伜.
  • Sulfate (SO鈧劼测伝) salts are generally soluble, but there are exceptions including Sr虏鈦, Ba虏鈦, Pb虏鈦, and Ca虏鈦.
  • Most sulfide (S虏鈦), carbonate (CO鈧兟测伝), phosphate (PO鈧劼斥伝), and hydroxide (OH鈦) salts are only slightly soluble, except when paired with Na鈦, K鈦, or NH鈧勨伜.
By applying these rules, you can often predict the formation of a precipitate in a reaction. Knowing which substances dissolve and which don't is key to writing accurate net ionic equations.
Chemical Equation Balancing
Balancing chemical equations is a fundamental skill required for solving any reaction equation. Balancing ensures that the same number of each type of atom appears on both sides of the equation, adhering to the Law of Conservation of Mass. This principle states that mass cannot be created or destroyed in a closed system, so everything you start with in a reaction must still be around at the end.

To balance a chemical equation, follow these steps:
  • Write the unbalanced equation with the correct formulae for reactants and products.
  • List and count all the atoms of each element in reactants and products.
  • Adjust coefficients 鈥 the numbers in front of compounds 鈥 to ensure the same number of each type of atom on both sides of the equation.
  • Check the equation to make sure all atoms balance and that the coefficients are in the simplest possible ratio.
An adequately balanced equation reflects the real stoichiometry of a chemical reaction, and without it, you wouldn't be able to accurately calculate concentrations or yields. This process not only aids in problem-solving but also deepens your understanding of how chemical reactions occur.

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Most popular questions from this chapter

The equation for the reaction of phosphorus and chlorine is \(\mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) .\) If you use 8000 molecules of \(P_{4}\) in this reaction how many molecules of \(\mathrm{Cl}_{2}\) are required to consume the \(\mathrm{P}_{4}\) completely?

The following compounds are water-soluble. What ions are produced by each compound in aqueous solution? (a) KOH (b) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{LiNO}_{3}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

In the following reactions, decide which reactant is oxidized and which is reduced. Designate the oxidizing agent and the reducing agent. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+3 \mathrm{Sn}^{2+}(\mathrm{aq})+14 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Sn}^{4+}(\mathrm{aq})+21 \mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{FeS}(\mathrm{s})+3 \mathrm{NO}_{3}^{-}(\mathrm{aq})+4 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{NO}(\mathrm{g})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Write balanced chemical equations for the following reactions: (a) production of ammonia, \(\mathrm{NH}_{3}(\mathrm{g}),\) by combin\(\operatorname{ing} \mathrm{N}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (b) production of methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\ell)\) by combining \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) (c) production of sulfuric acid by combining sulfur, oxygen, and water

Determine the oxidation number of each element in the following ions or compounds. (a) \(\mathrm{PF}_{6}\) (b) \(\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\) (c) \(\mathrm{UO}^{2+}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{POCl}_{3}\) (f) \(\mathrm{XeO}_{4}^{2-}\)

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