11. Consider The Following Reaction.$\[ Na_2CO_3(aq) + Mg(NO_3)_2(aq) \rightarrow MgCO_3(s) + 2NaNO_3(aq) \\]a. Calculate The Mass Of $\[ MgCO_3 \\] That Can Be Precipitated By Mixing 10.0 ML Of A $\[ 0.200 \, M \, Na_2CO_3

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11. Precipitation Reaction: Calculating the Mass of MgCO3

Understanding the Reaction

The given reaction involves the precipitation of magnesium carbonate (MgCO3) from a solution of sodium carbonate (Na2CO3) and magnesium nitrate (Mg(NO3)2). The reaction is as follows:

Na2CO3(aq)+Mg(NO3)2(aq)→MgCO3(s)+2NaNO3(aq){ Na_2CO_3(aq) + Mg(NO_3)_2(aq) \rightarrow MgCO_3(s) + 2NaNO_3(aq) }

In this reaction, the sodium carbonate (Na2CO3) acts as a precipitating agent, reacting with magnesium nitrate (Mg(NO3)2) to form magnesium carbonate (MgCO3) and sodium nitrate (NaNO3).

Calculating the Mass of MgCO3

To calculate the mass of MgCO3 that can be precipitated, we need to determine the number of moles of MgCO3 formed. We can do this by first calculating the number of moles of Na2CO3 present in the solution.

Step 1: Calculate the Number of Moles of Na2CO3

The number of moles of Na2CO3 can be calculated using the formula:

Number of moles=Molarity×Volume (in liters){ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} }

Given that the molarity of Na2CO3 is 0.200 M and the volume is 10.0 mL (or 0.0100 L), we can calculate the number of moles as follows:

Number of moles of Na2CO3=0.200 M×0.0100 L=0.00200 mol{ \text{Number of moles of Na}_2\text{CO}_3 = 0.200 \, \text{M} \times 0.0100 \, \text{L} = 0.00200 \, \text{mol} }

Step 2: Determine the Molar Ratio of Na2CO3 to MgCO3

From the balanced equation, we can see that 1 mole of Na2CO3 reacts with 1 mole of Mg(NO3)2 to form 1 mole of MgCO3. Therefore, the molar ratio of Na2CO3 to MgCO3 is 1:1.

Step 3: Calculate the Number of Moles of MgCO3

Since the molar ratio of Na2CO3 to MgCO3 is 1:1, the number of moles of MgCO3 formed is equal to the number of moles of Na2CO3 present in the solution.

Number of moles of MgCO3=0.00200 mol{ \text{Number of moles of MgCO}_3 = 0.00200 \, \text{mol} }

Step 4: Calculate the Mass of MgCO3

The molar mass of MgCO3 is 84.31 g/mol. We can calculate the mass of MgCO3 formed by multiplying the number of moles by the molar mass:

Mass of MgCO3=Number of moles×Molar mass{ \text{Mass of MgCO}_3 = \text{Number of moles} \times \text{Molar mass} }

Mass of MgCO3=0.00200 mol×84.31 g/mol=0.1696 g{ \text{Mass of MgCO}_3 = 0.00200 \, \text{mol} \times 84.31 \, \text{g/mol} = 0.1696 \, \text{g} }

Therefore, the mass of MgCO3 that can be precipitated by mixing 10.0 mL of a 0.200 M Na2CO3 solution is 0.1696 g.

Conclusion

In this problem, we calculated the mass of MgCO3 that can be precipitated by mixing 10.0 mL of a 0.200 M Na2CO3 solution. We first calculated the number of moles of Na2CO3 present in the solution, then determined the molar ratio of Na2CO3 to MgCO3, and finally calculated the mass of MgCO3 formed. The result shows that the mass of MgCO3 that can be precipitated is 0.1696 g.

Key Takeaways

  • The molar ratio of Na2CO3 to MgCO3 is 1:1.
  • The number of moles of MgCO3 formed is equal to the number of moles of Na2CO3 present in the solution.
  • The mass of MgCO3 formed can be calculated by multiplying the number of moles by the molar mass.

Practice Problems

  1. Calculate the mass of MgCO3 that can be precipitated by mixing 20.0 mL of a 0.100 M Na2CO3 solution.
  2. Determine the molar ratio of Na2CO3 to MgCO3 in the reaction.
  3. Calculate the number of moles of MgCO3 formed when 10.0 mL of a 0.200 M Na2CO3 solution is mixed with 10.0 mL of a 0.100 M Mg(NO3)2 solution.
    Q&A: Precipitation Reaction and Calculating the Mass of MgCO3

Frequently Asked Questions

In this article, we will address some of the most common questions related to the precipitation reaction and calculating the mass of MgCO3.

Q: What is the molar ratio of Na2CO3 to MgCO3 in the reaction?

A: The molar ratio of Na2CO3 to MgCO3 in the reaction is 1:1. This means that 1 mole of Na2CO3 reacts with 1 mole of Mg(NO3)2 to form 1 mole of MgCO3.

Q: How do I calculate the number of moles of MgCO3 formed?

A: To calculate the number of moles of MgCO3 formed, you need to determine the number of moles of Na2CO3 present in the solution. You can do this by multiplying the molarity of Na2CO3 by the volume of the solution in liters. The number of moles of MgCO3 formed is equal to the number of moles of Na2CO3 present in the solution.

Q: What is the molar mass of MgCO3?

A: The molar mass of MgCO3 is 84.31 g/mol. This value can be used to calculate the mass of MgCO3 formed by multiplying the number of moles by the molar mass.

Q: How do I calculate the mass of MgCO3 formed?

A: To calculate the mass of MgCO3 formed, you need to multiply the number of moles of MgCO3 by the molar mass of MgCO3. The formula for this calculation is:

Mass of MgCO3=Number of moles×Molar mass{ \text{Mass of MgCO}_3 = \text{Number of moles} \times \text{Molar mass} }

Q: What are some common mistakes to avoid when calculating the mass of MgCO3?

A: Some common mistakes to avoid when calculating the mass of MgCO3 include:

  • Not converting the volume of the solution from milliliters to liters
  • Not multiplying the molarity by the volume to get the number of moles
  • Not using the correct molar mass of MgCO3
  • Not performing the calculation correctly

Q: Can I use this method to calculate the mass of MgCO3 formed in any precipitation reaction?

A: No, this method is specific to the reaction between Na2CO3 and Mg(NO3)2. The molar ratio and molar masses of the reactants and products may be different in other precipitation reactions, so you will need to use a different method to calculate the mass of the product.

Q: What are some real-world applications of precipitation reactions?

A: Precipitation reactions have many real-world applications, including:

  • Water treatment: Precipitation reactions can be used to remove impurities from water by forming insoluble compounds that can be easily removed.
  • Industrial processes: Precipitation reactions can be used to produce a wide range of products, including pigments, dyes, and other chemicals.
  • Environmental remediation: Precipitation reactions can be used to remove pollutants from the environment by forming insoluble compounds that can be easily removed.

Conclusion

In this article, we have addressed some of the most common questions related to the precipitation reaction and calculating the mass of MgCO3. We have provided step-by-step instructions for calculating the mass of MgCO3 formed and have highlighted some common mistakes to avoid. We have also discussed some real-world applications of precipitation reactions and have provided a list of frequently asked questions and answers.