How many moles of iron are needed to produce 8.2mole of silver?

Answer 1

This question needs more information to solvable. Let's suppose a single displacement reaction like iron metal reacting with aqueous silver nitrate to produce aqueous iron(II) nitrate and silver metal. I don't know if this reaction answers your specific inquiry but it will serve to show how this type of calculation can be completed.

The unbalanced chemical equation would be: Fe + AgNO₃ --> Fe(NO₃)₂ + Ag . To balance this equation, follow these steps:

1. list the elements on each side
2. count the number of atoms of each element
3. add a coefficient in front of one substance to correct an imbalance (my recommendation is to find something that needs to be doubled and add a coefficient of 2 in front of it)
4. update the atom count.
5. If needed repeat steps 3 and 4 until both sides are balanced.

The balanced equation would be: Fe + 2AgNO₃ --> Fe(NO₃)₂ + 2 Ag

We are given that n_Ag = 8.2 mol. The mole to mole ratio of iron to silver in this equation is 1:2

You would set up your factor label so that 8.2 mol Ag is multiplied by 1 mol Fe and then divided by 2 mol of Ag to equal 4.1 mol of Fe.

This is an oxidation reduction reaction which ordinarily needs to be balanced in a special manner. However, balancing the way outlined above and balancing it by the method of half reactions produce the same balanced equation.