[Dt_rox]

I know how to do these kinds of double displacement reactions, but I have no clue where the water goes, can anyone help me out on this???

It would normaly go like this without the hydrate I think:

Na2CO3 (aq) + Ca(NO3)2 (aq) --> 2NaNO3(aq) + CaCO3(s)

So where would the 4H2O go if the reaction was like this:

Na2CO3 (aq) + Ca(NO3)2*4H2O (aq) --> ?

[Borek]

These reactions go in the solution, so hydration water is already lost from the particle.

[Donaldson Tan]

Consider it as a 2-step process:
1. Ca(NO3)2.4H2O (s) => Ca(NO3)2 (aq) + 4H2O (l)
2. Na2CO3 (aq) + Ca(NO3)2 (aq) --> 2NaNO3(aq) + CaCO3(s)

The water of crystallisation becomes part of the solvent.

[Dt_rox]

Ok, now that i know what happens, tell me if this is right:

0.02L of .175 mol/L of Na2CO3

cv=n
.02 x .175 = n
0.0035 mol = n

0.0035 mol x 106 = 0.371g (dissolved in water)

0.025L of 0.125 mol/L of Ca
cv=n
0.025 x 0.125 = n
0.003125mol = n

0.003125 x 236.15 = 0.738g (then poured in water)

I dont feel like typing out the limiting reactant thing, so i will just tell you guys that its Ca(NO3).4H2O.

So since the precipitate will be CaCO3, and it will have 0.003125 moles, do i just multiply it by the molar mass of CaCO3 to get 0.313g and just leave out the water? Because the question asks to find the mass of the precipitate. Thanks.

[Dt_rox]

Oh one more thing, is this the full equation?

Na2CO3(aq) + Ca(NO3)2.4H2O (aq)--> 2NaNO3(aq) + CaCO3(s) + 4H2O

or...

Na2CO3(aq) + Ca(NO3)2(aq) --> 2NaNO3(aq) + CaCO3(s)

[Donaldson Tan]

Na2CO3(aq) + Ca(NO3)2.4H2O (aq)--> 2NaNO3(aq) + CaCO3(s) + 4H2O

The state should be (s) to indicate Calcium Nitrate is solid initally.

[Dt_rox]

The state should be (s) to indicate Calcium Nitrate is solid initally.

But the Ca(NO3)2 is dissolved in water to make a solution, and when the two solutions mix is when the precipitate is formed, so shouldnt every reactant be aqueous, and only the precipitate be solid? Because the Na2CO3 was solid initially too, but then was dissolved in water after.