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

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.