anonymous
  • anonymous
The minerals in seawater can be obtained through evaporation. For every liter of seawater evaporated, 3.7 grams of MgOH can be obtained. How many liters of seawater must be evaporated to collect 5.00 moles of MgOH?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
convert 3.7 grams of MgOH to moles, then do multiplication. OTL
anonymous
  • anonymous
what would i multiply it by tho?
anonymous
  • anonymous
First we need to get number of moles of MgOH in a liter of water. Molecular Weight of MgOH = Weight of Magnesium (Unfortunately I can't recall it) + 16 (Weight of Oxygen) + 1 (Weight of Hydrogen) \[\mathaf{ \text{Moles of MgOH in a liter of water} = \frac{3.7}/{\text{Molecular Weight}}}\]

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anonymous
  • anonymous
no. liter of water is not needed. just convert everything to moles and divide the amount you want by the amount produced. that gives # of liters. OTL. btw it's 1 mile ~ 22.4 L, don't forgot to convert there as well.
anonymous
  • anonymous
*mole
anonymous
  • anonymous
Sorry for the LaTeX, somehow my preview of the post isn't working.\[\mathsf{ \text{Moles of MgOH in a liter of water} = \frac{3.7}{\text{Molecular Weight}}}\]
anonymous
  • anonymous
the answer is 78.4L anyone know how i can get there?
anonymous
  • anonymous
Oh yeah, I forgot to divide it by 22.4. After you get moles in a liter then you can divide 5 by the moles of MgOH in a liter of water and the number you get is the number of liter of water you need to evaporate to get 5 Moles of MgOH. I am sorry for the delay, my preview isn't working.
anonymous
  • anonymous
Or, you can follow RONCC do it in much simpler way then mine
JFraser
  • JFraser
the formula is Mg(OH)2, not MgOH. Your molar mass will have to be recalculated. 22.4L is the volume of 1mol of GAS, not LIQUID. Find the molar mass of Mg(OH)2. Find the mass of 5 moles of Mg(OH)2. Find how many times 3.7g fits into that mass. There's your answer \[(5.0mol Mg(OH){_2}) * (\frac{58.3g Mg(OH){_2}}{1mol Mg(OH){_2}}) * (\frac{1L water}{3.7g Mg(OH){_2}}) = 78.8L\]

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