/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 If solutions of iron (III) nitra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If solutions of iron (III) nitrate and sodium carbonate are mixed, what would be the formula of the precipitate? (A) \(\mathrm{Fe} 3 \mathrm{CO}_{3}\) (B) \(\mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}\) (C) \(\mathrm{NaNO}_{3}\) (D) No precipitate would form.

Short Answer

Expert verified
The formula of the precipitate that forms when iron(III) nitrate and sodium carbonate are mixed is \(\mathrm{Fe({CO_{3})_{3}}\).

Step by step solution

01

Write Down the Chemical Reaction

First write down the reaction: \(\mathrm{Fe(NO_{3})_{3}}\) + \(\mathrm{3Na_{2}CO_{3}}\) -> \(\mathrm{2Fe({CO_{3})_{3}}\) + \(\mathrm{6NaNO_{3}}\) Here, iron(III) nitrate \(\mathrm{Fe(NO_{3})_{3}}\) and sodium carbonate \(\mathrm{Na_{2}CO_{3}}\) are the reactants. A possible product is iron carbonate \(\mathrm{Fe({CO_{3})_{3}}\) and sodium nitrate \(\mathrm{NaNO_{3}}\).
02

Check the Solubility

The next step is to check the solubility of the products. Sodium nitrate, \(\mathrm{NaNO_{3}}\), is soluble in water, so it stays in solution. The solubility rules say that most compounds with carbonate, \(\mathrm{CO_{3}^{2-}}\), are insoluble. This means, that iron(III) carbonate, \(\mathrm{Fe({CO_{3})_{3}}\), would precipitate, as this compound is insoluble in water.
03

Write the Formula of the Precipitate

Since iron(III) carbonate is the compound that would precipitate, its formula \(\mathrm{Fe({CO_{3})_{3}}\) is the answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Precipitate Formation
When observing chemical reactions, especially those occurring in aqueous solutions, one common phenomenon is the formation of a precipitate. A precipitate is a solid that forms when two solutions are mixed and undergo a chemical reaction. This solid emerges because its components are more stable as a solid rather than staying dissolved in the solution.
An important aspect of precipitate formation is recognizing that it signifies a reaction has taken place. This offers chemists a visual cue that substances have interacted to create a new product. For example, when mixing solutions of iron(III) nitrate and sodium carbonate, iron(III) carbonate forms as a precipitate due to its insolubility in water. This solid compound visibly separates from the aqueous solution, showcasing a successful chemical transformation.
Solubility Rules
Solubility rules are a set of guidelines that help predict whether a compound will dissolve in water. These rules are crucial for identifying if a compound will stay dissolved or form a precipitate in a reaction. Generally, compounds containing alkali metals (like sodium or potassium) and nitrates are soluble in water.
However, most carbonates are insoluble, especially when combined with transition metals such as iron. In our example with iron(III) nitrate and sodium carbonate, solubility rules indicate that while sodium nitrate stays dissolved, iron(III) carbonate does not. Therefore, the insoluble nature of iron(III) carbonate leads to its appearance as a precipitate, providing clarity and direction for identifying products in such chemical reactions.
Iron(III) Carbonate
Iron(III) carbonate is a chemical compound, often represented by the formula \(Fe_2(CO_3)_3\). This compound is known for its low solubility in water, making it a prime candidate for precipitate formation in aqueous reactions. When solutions containing iron(III) ions and carbonate ions are mixed, these ions quickly combine due to their chemical compatibility.
This compatibility stems from the need for iron(III) ions to neutralize their positive charge with the negative charge of carbonate ions. As a result, they form iron(III) carbonate, a solid compound that precipitates out of solution. Understanding this process not only highlights the chemistry behind precipitate formation but also reinforces solubility rules, effectively linking theory to practical observation.
Chemical Equations
Chemical equations are a vital tool in representing chemical reactions. They display the reactants, products, and their coefficients, providing a clear understanding of how substances interact. Writing a balanced chemical equation is crucial because it reflects the conservation of mass and atoms.
In our example, the equation \(Fe(NO_3)_3 + 3Na_2CO_3 \to 2Fe_2(CO_3)_3 + 6NaNO_3\) not only shows the reactants and potential products but also highlights the stoichiometry of the reaction. Here, the coefficients ensure that the number of each atom on the reactant side matches that on the product side, preserving mass balance in the equation. By understanding chemical equations, one can predict reaction outcomes and the formation of products like precipitates, providing a comprehensive view of chemical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) What is the enthalpy change for reaction 1\(?\) (A) \(-148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (B) \(-56 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (C) \(-18 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (D) \(+148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\)

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. Identify the correct net ionic equation for the reaction that takes place. (A) \(\mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q) \mapsto \mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (B) \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \leftrightarrow \mathrm{NaCN}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (D) \(\mathrm{H}^{+}(a q)+\mathrm{CN}^{-}(a q)+\mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q)+\mathrm{Na}^{+}\) (aq)

\(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) If the reaction above took place at standard temperature and pressure and 150 grams of \(\mathrm{CaCO}_{3}(\mathrm{s})\) were consumed, what was the volume of \(\mathrm{CO}_{2}(g)\) produced at STP? (A) 11 L (B) 22 L (C) 34 L (D) 45 L

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.