Molarity

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  • Oh hey Skylar, I cannot figure out this molarity problem.
  • Hey Dakota, what is wrong?
  • Well the problem is to find the ion's molarity of 150 mL of 5.35M of Al(NO3)3 and 250mL of Li2CrO4
  • Really? I can help you with it. What is the problem?
  • 5.35 M Al(NO3)3 = x/.15L = .803 moles Al(NO3)3
  • 7.35 M Li2CrO4 = x/.25L = 1.84 moles Li2CrO4
  • Well you start by finding the moles of Li2CrO4 and Al(No3)3. You do that by setting up the molarity equation but have moles as x and changing mL into L. Look at the board.
  • .803 mol Al(NO3)3 x 3mol No3/ 1 mol Al(NO3)3 = 2.41 mol/.3L = 8.03 M NO3
  • 1.84 mol Li2CrO4 x 2 mol Li / 1 mol Li2CrO4 = 3.68 mol Li / .5 = 7.36M Li
  • After you find the Moles of each formula then we find the molarity of each ion by stoiching and dividing each ion's moles by double of the liters.
  • What??
  • 0M = Al
  • Aluminum, however, has zero molarity because it is part of the limiting react, Al(NO3)3, and the solid, Al2(CrO4)3
  • .803 - 1.84 = 1.04mol x 3mol CrO4 / 1 mole CrO4 = 3.12 mole CrO4 / .4L = 7.80M CrO4
  • I GET IT! THANK YOU DAKOTA!
  • In order to find, lithium, we must take the moles of both reactions and subtract them together, stoich the difference and divide the result by the sum of both liters and that is how you do the problem.
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