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Chapter 4: Problem 90

A flask contains \(49.8 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) (calcium hydroxide). How many milliliters of \(0.350 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (sodium carbonate) are required to react completely with the calcium hydroxide in the following reaction? $$ \mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{CaCO}_{3}(s)+2 \mathrm{NaOH}(a q) $$

Short Answer

Expert verified
21.34 mL of sodium carbonate solution is required.

Step by step solution

01

Calculate Moles of Calcium Hydroxide

Use the molarity and the volume of calcium hydroxide to determine the number of moles. The formula for moles is \( ext{moles} = ext{molarity} imes ext{volume in liters} \). For \( ext{volume in liters} \): \( 49.8 ext{ mL} = 0.0498 ext{ L} \).\[ \text{moles of } \mathrm{Ca(OH)}_2 = 0.150 ext{ M} \times 0.0498 ext{ L} = 0.00747 ext{ moles} \]
02

Determine Moles of Sodium Carbonate Required

Based on the balanced chemical equation, 1 mole of \( \mathrm{Na}_2\mathrm{CO}_3 \) reacts with 1 mole of \( \mathrm{Ca(OH)}_2 \). Thus, you will need 0.00747 moles of \( \mathrm{Na}_2\mathrm{CO}_3 \) to react completely with the calcium hydroxide from Step 1.
03

Calculate Volume of Sodium Carbonate Solution Needed

Now, use the number of moles of \( \mathrm{Na}_2\mathrm{CO}_3 \) and its molarity to find the required volume. Using the formula \( ext{volume} = \frac{\text{moles}}{\text{molarity}} \), calculate:\[ \text{volume of } \mathrm{Na}_2\mathrm{CO}_3 = \frac{0.00747 \text{ moles}}{0.350 \text{ M}} = 0.02134 \text{ L} \]Convert this volume into milliliters: \( 0.02134 \text{ L} = 21.34 \text{ mL} \).

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Key Concepts

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

Molarity and Its Role in Calculations
Molarity is a fundamental concept in chemistry that describes the concentration of a solution. It is defined as the number of moles of solute per liter of solution and is expressed in units of moles per liter (M). Understanding molarity is crucial because it helps chemists and students alike determine how much of a substance is present in a given volume of solution. To apply molarity in calculations:
  • Identify the molarity (concentration) of the solution from the problem.
  • Convert any necessary volume measurements from milliliters to liters, since molarity is defined per liter.
  • Use the formula: \[ \text{moles} = \text{molarity} \times \text{volume in liters} \]
In our exercise, the molarity of calcium hydroxide was 0.150 M, and the volume was converted from milliliters to liters. This conversion is essential for accurately calculating the number of moles involved in the reaction.
Understanding the Balanced Chemical Reaction
Chemical reactions involve the transformation of reactants into products. A balanced chemical equation reflects the conservation of mass and atoms during this process. It communicates the precise ratios in which substances react and are formed.In this exercise, the balanced chemical reaction is given by:\[ \mathrm{Na}_{2} \mathrm{CO}_{3}(aq) + \mathrm{Ca}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{CaCO}_{3}(s) + 2 \mathrm{NaOH}(aq) \]From this equation:
  • One mole of sodium carbonate reacts with one mole of calcium hydroxide.
  • Calcium carbonate and sodium hydroxide are formed as products.
This 1:1 stoichiometric relationship between sodium carbonate and calcium hydroxide means that the moles of sodium carbonate required will equal the moles of calcium hydroxide present in the solution, simplifying calculations.
Volume Calculations in Reactions
Calculating the volume of a solution required to fully react with another involves using the concept of molarity. After determining the number of moles of one reactant, you can find the volume of a second reactant at a known molarity needed to complete the reaction.Follow these steps in volume calculations:
  • Use the number of moles obtained from previous steps.
  • Apply the formula: \[ \text{volume} = \frac{\text{moles}}{\text{molarity}} \]
  • Convert the resultant volume from liters to milliliters if necessary by multiplying by 1000.
For sodium carbonate, we calculated the volume needed using its molarity and the moles required to react with calcium hydroxide. The result was 0.02134 liters, which is equivalent to 21.34 milliliters, ensuring the complete reaction in the specified conditions.

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Most popular questions from this chapter

A sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), weighing \(1.192 \mathrm{~g}\) is placed in a \(100.0-\mathrm{mL}\) volumetric flask, which is then filled to the mark with water. What is the molarity of the solution?

A metal, \(\mathrm{M}\), was converted to the sulfate, \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). Then a solution of the sulfate was treated with barium chloride to give barium sulfate crystals, which were filtered off. \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{BaCl}_{2}(a q) \longrightarrow 2 \mathrm{MCl}_{3}(a q)+3 \mathrm{BaSO}_{4}(s)\) If \(1.200 \mathrm{~g}\) of the metal gave \(6.026 \mathrm{~g}\) of barium sulfate, what is the atomic weight of the metal? What is the metal?

An antacid tablet has calcium carbonate as the active ingredient; other ingredients include a starch binder. You dissolve the tablet in hydrochloric acid and filter off insoluble material. You add potassium oxalate to the filtrate (containing calcium ion) to precipitate calcium oxalate. If a tablet weighing \(0.680 \mathrm{~g}\) gave \(0.629 \mathrm{~g}\) of calcium oxalate, what is the mass percentage of active ingredient in the tablet?

Iron(III) chloride can be prepared by reacting iron metal with chlorine. What is the balanced equation for this reaction? How many grams of iron are required to make \(3.00 \mathrm{~L}\) of aqueous solution containing \(9.00 \%\) iron(III) chloride by mass? The density of the solution is \(1.067 \mathrm{~g} / \mathrm{mL}\).

Balance the following oxidation-reduction reactions by the half-reaction method. a. \(\mathrm{CuCl}_{2}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{Cu}(s)\) b. \(\mathrm{Cr}^{3+}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Cr}(s)+\mathrm{Zn}^{2+}(a q)\)
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