Chapter 4: Problem 38
What mass of \(\mathrm{MgNH}_{4} \mathrm{PO}_{4}\) precipitated when \(200.0 \mathrm{~mL}\) of a \(1.000 \%(w / v)\) solution of \(\mathrm{MgCl}_{2}\) were treated with \(40.0 \mathrm{~mL}\) of \(0.1753 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and an excess of \(\mathrm{NH}_{4}^{+}\)? What was the molar concentration of the excess reagent \(\left(\mathrm{Na}_{3} \mathrm{PO}_{4}\right.\) or \(\left.\mathrm{MgCl}_{2}\right)\) after the precipitation was complete?
Short Answer
Step by step solution
Determine Moles of MgClâ‚‚
Determine Moles of Na₃PO₄
Write the Precipitation Reaction
Identify Limiting Reactant
Calculate Moles of MgNHâ‚„POâ‚„ Formed
Calculate Mass of MgNHâ‚„POâ‚„
Calculate Concentration of Excess MgClâ‚‚
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.
Precipitation Reaction
- Precipitation reactions are vital for simple salt determinations.
- They involve combining solutions to form a solid product.
- The formed solid can be separated by filtration for further analysis.
Limiting Reactant
- Calculate the number of moles of each reactant available.
- Use the stoichiometry of the balanced equation to compare the mole ratio present to the mole ratio needed.
- The reactant with fewer moles than needed is the limiting reactant.
Molar Mass Calculation
- \( \text{Molar mass of } \text{MgNH}_4\text{PO}_4 = 24.31 \, + \, 14.01 \, + \, 30.97 \, + \, (16.00 \, \times 4) \, = \) 137.29 g/mol.
Molar Concentration
- Determine the number of moles of solute present.
- Measure the volume of the solution in liters.
- Divide the moles of solute by the volume of the solution in liters.