Fe(NO3)3 + 3NaOH ---> Fe(OH)3 + 3Na(NO)3

Question

a. Given 7.45 mol of Fe(NO3)3 with an excess of 3NaOH, calculate the mol of 3Na(NO)3 that should be formed

b. Given 7.45 g of Fe(NO3)3 with an excess of 3NaOH, calculate the mass of 3Na(NO)3 that should be formed

Answer 1

For part a, from the balanced equation you know that one mol of Fe(NO₃)₃ will produce 3 moles of Na(NO)₃.

Using ratios, you can find that 7.45 moles of Fe(NO₃)₃ will produce (7.45 * 3) moles of Na(NO)₃ = 22.4 moles.

For part b, first convert grams into moles: 7.45 g / 242 (g/mol) = 0.0308 moles of Fe(NO₃)₃. And like part a, you can calculate the moles of Na(NO)₃ produced by multiplying by 3. The you can convert back to grams by multiplying by the molecular weight of Na(NO)₃.