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.

2 Answers

Relevance
  • 4 months ago
    Favorite 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 disssociates

    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.

    • Login to reply the answers
  • 4 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.

    • Login to reply the answers
Still have questions? Get your answers by asking now.