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Calculate the mass of \(\mathrm{MgCO}_{3}\) precipitated by mixing \(10.0 \mathrm{mL}\) of a \(0.200 M \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution with \(5.00 \mathrm{mL}\) of \(0.0500 M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

Short Answer

Expert verified
Answer: The mass of MgCO₃ precipitated is 0.0211 grams.

Step by step solution

01

Write a balanced chemical equation

First, let's write a balanced chemical equation for this reaction. Sodium carbonate reacts with magnesium nitrate to form magnesium carbonate and sodium nitrate: \(\mathrm{Na}_{2}\mathrm{CO}_{3}(aq) + \mathrm{Mg}(\mathrm{NO}_{3})_{2}(aq) \rightarrow \mathrm{MgCO}_{3}(s) + 2\mathrm{NaNO}_{3}(aq)\) This reaction is balanced.
02

Determine the limiting reagent

To determine the limiting reagent, we need to calculate the number of moles of both reactants and compare their mole ratios: Moles of Na2CO3 = volume (L) * molarity Moles of Na2CO3 = 0.0100 L * 0.200 mol/L = 0.00200 mol Moles of Mg(NO3)2 = volume (L) * molarity Moles of Mg(NO3)2 = 0.00500 L * 0.0500 mol/L = 0.000250 mol Next, compare the mole ratios: Mole ratio of Na2CO3 to Mg(NO3)2 = (0.00200 mol Na2CO3) / (0.000250 mol Mg(NO3)2) = 8.00 Based on the balanced equation, the mole ratio required for Na2CO3 to Mg(NO3)2 is 1:1. However, the actual mole ratio is 8:1. This means that Mg(NO3)2 is the limiting reagent.
03

Calculate the moles of MgCO3 formed

Since Mg(NO3)2 is the limiting reagent, we can use its moles to find the moles of MgCO3 produced: Moles of MgCO3 = Moles of Mg(NO3)2 (1:1 ratio) Moles of MgCO3 = 0.000250 mol
04

Calculate the mass of MgCO3 precipitated

Using the moles of MgCO3 formed and the molar mass of MgCO3, we can find the mass of MgCO3 precipitated: Mass of MgCO3 = Moles of MgCO3 * Molar mass of MgCO3 Mass of MgCO3 = 0.000250 mol * 84.31 g/mol = 0.0211 g Therefore, 0.0211 grams of MgCO3 are precipitated by mixing the two solutions.

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Key Concepts

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

Limiting Reagent
In a chemical reaction, the limiting reagent is the reactant that gets entirely consumed first, halting further reaction progress. It determines the maximum amount of product that can be generated. Imagine baking cookies: if you have less flour than needed for a batch, your flour will limit how many cookies you can make, no matter how much of other ingredients you have.
In the reaction between sodium carbonate and magnesium nitrate, the limiting reagent is necessary to identify to calculate the product amount. To find it, we count the moles of all reactants. Here, we had:
  • 0.00200 mol of Na2CO3
  • 0.000250 mol of Mg(NO3)2
From the balanced equation, we see a 1:1 mole ratio is needed between Na2CO3 and Mg(NO3)2. Given our quantities, Mg(NO3)2 is less than Na2CO3, indicating it will run out first, making it our limiting reagent. Always remember: reactions stop once the limiting reagent is used up, which in this case, limits how much MgCO3 can be formed.
Mole Calculations
Mole calculations are essential for quantifying substances used and formed in a chemical reaction. The mole is a fundamental unit in chemistry that counts entities like atoms, molecules, ions, etc. Think "dozens" but for chemistry, with 1 mole equaling approximately 6.022 x 10^23 entities.
When it comes to preparing or analyzing reactions, we often start with concentrations given in moles per liter (mol/L) and volumes in liters. By multiplying concentration by volume, you directly calculate moles. For instance, in our exercise:
  • For Na2CO3: 0.0100 L * 0.200 mol/L = 0.00200 mol
  • For Mg(NO3)2: 0.00500 L * 0.0500 mol/L = 0.000250 mol
Understanding these mole calculations is the first step in much of chemical problem-solving, providing a quantitative backbone to our reactions.
Precipitation Reactions
Precipitation reactions result in the formation of a solid from a solution. When two aqueous solutions are mixed, the ions in them can sometimes combine to form an insoluble ionic compound, known as a precipitate. Imagine mixing two clear liquids and witnessing a solid powdery substance appearing from nowhere.
This solid formation occurs due to the product's low solubility in the solution. In the example of Na2CO3 and Mg(NO3)2, magnesium carbonate (MgCO3) precipitates because it doesn't remain dissolved in water easily. The balanced equation: Na2CO3(aq) + Mg(NO3)2(aq) → MgCO3(s) + 2NaNO3(aq) shows this, where MgCO3 is the precipitate, denoted by the (s) for solid.
Precipitation reactions are fundamental for tasks such as removing ions from solutions in water treatment or in lab settings for illustrating chemical principles. Spotting these reactions and understanding the conditions under which they occur is vitally useful in predicting outcomes in chemistry.

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Most popular questions from this chapter

Superoxide Dismutases Oxygen in the form of superoxide ions, \(\mathrm{O}_{2}^{-}\), is quite hazardous to human health. Superoxide dismutases represent a class of enzymes that convert superoxide ions to hydrogen peroxide and oxygen by the unbalanced chemical equation: $$ \mathrm{O}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{O}_{2}(a q) $$ a. Identify the oxidation and reduction half-reactions. b. Balance the equation.

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