Finding volume of one compound using Stoichiometry equation

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SUMMARY

The discussion focuses on calculating the volume of ammonium carbonate required to react with a given volume of silver (I) nitrate using stoichiometry. The balanced chemical equation provided is 2AgNO3 (aq) + (NH4)2CO3 (aq) --> AgCO3 (s) + 2NH4NO3 (aq). The user successfully calculated the moles of silver (I) nitrate as 0.3 mol and determined the moles of ammonium carbonate needed as 0.15 mol. However, the lack of molarity for the ammonium carbonate solution prevents further calculation of its volume.

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  • Understanding of stoichiometry and balanced chemical equations
  • Knowledge of molarity calculations
  • Familiarity with the concept of moles in chemistry
  • Ability to manipulate equations involving concentration and volume
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  • Research how to determine the molarity of a solution from mass and volume
  • Learn about stoichiometric calculations in chemical reactions
  • Explore the concept of limiting reagents in chemical equations
  • Study the preparation of ammonium carbonate solutions and their concentrations
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Homework Statement


I am having a problem finding the volume of one compound while having the molarity and volume for another using a balanced equation.
This is the problem statement: What volume in mL of ammonium carbonate is needed to react completely with 200.0 mL of 1.500 M silver (I) nitrate to the following equation?
2AgNO3 (aq) + (NH4)2CO3 (aq) --> AgCO3 (s) + 2NH4NO3 (aq)

Homework Equations


M1 V1 x M2 V2

The Attempt at a Solution


I am able to convert the silver (I) nitrate using the molarity and volume values to find its moles by using this equation 1.500 M = moles / 0.200 L => moles = 0.3 mol AgNO3. Next, I used stoichiometry to find the moles of ammonium carbonate with this equation 0.3 mol AgNO3 x (1 mol (NH4)2CO3/2 mol AgNO3) = 0.15 mol (NH4)CO3. Can someone help me figure out how to get the volume if I do not have the molarity of ammonium carbonate?
 
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I think you need the molarity of the ammonium carbonate solution. Without it, there is insufficient information to work the problem. (And from a quick glance at it, what you did is correct).
 
As Charles signaled, no way to answer the problem with the information given.
 
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