<span> first calculate the moles of every compound used
n NH3 = (0.160 mol /L )(0 .360 L) = 0.058 moles NH3
which is equal to moles NH4OH
0.058 moles NH4OH.
n MgCl2 = (0.12 mol /L)( 0.10 liters = .012 moles MgCl2
reaction 2 NH3.H20 or 2 NH4OH + MgCl2 --> Mg(OH)2(s) + 2 NH4+ + 2Cl-
Molarity of Mg++ =(0.012 moles Mg+) / 0.46 liters = 0.026 Molar Mg++
Molarity OH= (0.058 moles OH-)/ 0.46 liters = 0.126 molar OH-
the limiting reactant is Mg ++ because it is lesser than the molarity of OH-
Now the challenge is to drive the OH-) concentration down so low that Mg(OH)2 will not precipitate out.
Ksp = 1.8 x 10^-11 = (Mg)(OH-)^2
Mg++ concentration to be .026 Molar, so let X = the (OH-) concentration
1.8 x l0^-11 = (0.026)(X)^2
(X)^2 = ( 1.8 x 10^-11) / (0.026)
X^2 = 6.92 x 10^-10
X =
2.63 x l0^-5 moles (OH-)/ L
this is the concentration where solid will form
so we need to lower the (OH-) which is
0.126 molar OH- down to 2.63 x 10^-5 molar OH- by adding NH4+ ions.
(0.126 moles/liter 0H-) - (2.63 x l0^-5 molar OH- ) = 0.123 moles per liter
OH- to be neutralized by adding NH4+
since the mole ratio is 1 : 1 then </span><span> 0.123 moles per liter NH4+ concentration to neutralize the </span><span><span> 0.123</span> moles of OH- in solution.
so to prevent the precipitation of mg(oh)2
0.123 - 0.058 = 0.065 Molar NH4+ is needed
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Explanation:
its answer is displacement reaction
Just by simply having a water temp.
I think it is gravitational pull
Hey there!
The molar mass of magnesium is 24.305 grams.
The molar mass of an element is the same as the atomic mass of an element, but measured in grams.
It's on the periodic table underneath the element symbol.
Hope this helps!
