A SOLUTION OF 1.000 L .750 M Na2CO3 mixed with 500.0 mL of a 1.50 M solution of Fe(NO3)3.?
predict the limiting reactant and calculate the moles of ions in excess.
so, I converted both compounds into moles and I got .750 moles for both.
Help, please & thank you.
- Trevor HLv 74 months agoFavorite Answer
The first step ( always) is to write a balanced equation for the reaction
2Fe(NO3)3(aq) + 3Na2CO3(aq) → Fe2(CO3)3(s) + 6NaNO3(aq)
2mol Fe(NO3)3react with 3 mol Na2CO3
You have calculate that you have 0.75 mol of each reactant in solution.
Now you have to determine the limiting reactant.
From the equation ,
0.75 mol Fe(NO3)3 will react with 0.75mol *3/2 = 1.125 mol Na2CO3
Because you have only 0.75 mol Na2CO3, this is not sufficient to react with all the Fe(NO3)3
Conclusion: The Na2CO3 is limiting
0.75 mol of Na2CO3 will react with 0.75 mol *2/3 = 0.500 mol Fe(NO3)3
You have 0.75 mol Fe(NO3)3
Therefore there will be 0.75 mol - 0.50 mol = 0.25 mol Fe(NO3)3 unreacted.
The question asks for mol of ions:
Na2CO3 → 2Na+ + CO3-
1mol → 3mol
1mol Na2CO3 will produce 3 mol ions
0.25 mol Na2CO3 will produce 0.25*3 = 0.75 mol ions unreacted.
- pisgahchemistLv 74 months ago
Mixing Na2CO3 and Fe(NO3)3 ....
The catch is that there will be NO Fe2(CO3)3 being formed. It's impossible. Iron(III) carbonate doesn't exist. A solution of Fe3+ ions is too acidic for Fe2(CO3)3 to be stable. Instead, the carbonate will go to CO2 gas. The actual reaction is:
3Na2CO3(aq) + 2Fe(NO3)3(aq) + 3H2O(l) --> 2Fe(OH)3(s) + 3CO2(g) + 6NaNO3(aq)
This means that your limiting reactant problem must take a different direction.