8 moles of fe(no3)3 reacts with Cu2(Fe(cn)6)to produce Cu(No3)2 and Fe4(Fe(CN)6)3
What is the mass in mg Cu(No3)2, can't get set up, calculated the molar mass 91 grams, can convert to mg, what am I not doing, trying to help my niece, I have not gone to school in a lot of years, can't see what step I am missing, should I be converting the 8 moles of Fe(NO3)3?
Thanks for your time

Duplicate

8 moles of fe(no3)3 reacts with Cu2(Fe(cn)6)to produce Cu(No3)2 and Fe4(Fe(CN)6)3
what is the mass in mg Cu(No3)2, can't get set up, calculated the molar mass 91 grams(NO3)3?


Fe(NO_3)_3\,+\,Cu_2(Fe(CN)_6) = Cu(NO_3)_2\,+\,Fe_4(Fe(CN)_6)_3

First you need to balance the equation so you know the molar ratios:

Fe(NO_3)_3\,+\,Cu_2(Fe(CN)_6) = 2Cu(NO_3)_2\,+\,Fe_4(Fe(CN)_6)_3

Think of the Fe(CN)_6^{-2} group as a discrete anion that doesn't change during the reaction (it might move around, but it remains the anion). That will simplify the balancing. Nitrate, NO_3^- is the same.

4Fe(NO_3)_3\,+\,3Cu_2A = 6Cu(NO_3)_2\,+\,Fe_4A_3

We now go back to the "full" formula by replacing A with Fe(CN)6. The balance formula is required to determine the number of moles of the other compounds that will be consumed or used.

4Fe(NO_3)_3\,+\,3Cu_2(Fe(CN)_6) = 6Cu(NO_3)_2\,+\,Fe_4(Fe(CN)_6)_3

We have 8 moles of Ferric Nitrate. Based on the balanced formula, that will require 6 moles of Cu2(Fe(CN)6) and will produce 12 moles of Cu(NO3)2 and 2 moles of Fe4(Fe(CN)6)3.

The molecular weight of Cu(NO3)2 is 187.57 g/mole (63.546+2*14.0067+6*15.9994. Since we're producing 12 moles of this material, the mass is

12 \cdot 187.57 = 2250.85\,grams

I think that's what you needed to calculate.