1. reaction of iron (ii) sulfate with sodium hydroxide
ion charges: (na+, fe2+, oh-, so42-)
write a balanced chemical equation
2. reaction of iron (iii) nitrate with sodium hydroxide
ion charges: (na+, fe3+, oh-, no3-)
write a balanced chemical equation
3.how many grams of sodium nitrate formed when 1.5 moles of iron (iii) nitrate reacted with excess of sodium hydroxide? show your work

1) FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4

2) Fe(NO3)3 + 3NaOH → Fe(OH)3 + 3NaNO3

3) There will be 4.5 moles of NaNO3 formed

Explanation:

1. Reaction of Iron (II) Sulfate with Sodium Hydroxide

Iron (II) Sulfate = FeSO4

Sodium Hydroxide = NaOH

FeSO4 + NaOH →  Fe^2+ + SO4^2- + Na+ +OH- → Fe(OH)2 + Na2SO4

On the right side we have 2x Na on the left side we have 1x Na so we have to multiply NaOH by 2

Now the equation is balanced and we have:

FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4

2. Reaction of Iron (III) Nitrate with Sodium Hydroxide

Iron (III) Nitrate = Fe(NO3)3

Sodium Hydroxide = NaOH

Fe(NO3)3 + NaOH → Fe^3+ + 3NO- + Na+ + OH-

Fe(NO3)3 +NaOH → Fe(OH)3 + NaNO3

On the right side we have 3x H on the left side 1x H so we have to multiply NaOH by 3

Fe(NO3)3 + 3NaOH → Fe(OH)3 + NaNO3

Now we have 3x Na on the left side, on the right side we have 1x Na

We have to multiply NaNO3 by 3

Fe(NO3)3 + 3NaOH → Fe(OH)3 + 3NaNO3

Now the equation is balanced

3.How many grams of sodium nitrate formed when 1.5 moles of iron (III) nitrate reacted with excess of sodium hydroxide?

Fe(NO3)3 + 3NaOH → Fe(OH)3 + 3NaNO3

For 1 mol of iron(III) nitrate we need 3 moles of sodium hydroxide to produce 1 mol of iron(III) hydroxide and 3 moles of sodium nitrate

For 1.5 moles of iron(III) nitrate we'll have 3*1.5 moles = 4.5 moles of sodium nitrate

There will be 4.5 moles of NaNO3 formed

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