Answer:
1. FeSO4+2NaOH→Na2SO4+Fe(OH)2
2. Fe(NO3)3+3NaOH→Fe(OH)3+3NaNO3
3. 382 g NaNO3
Explanation:
1. This is a double displacement reaction so perform exchanges with specimens and balance accordingly.
2. This is a double displacement reaction so perform exchanges with specimens and balance accordingly.
3. Start with 1.5 mol of Fe(NO3)3 then include mole to mole ratio of NaNO3 to Fe(NO3)3 then multiply by molar mass of Fe(NO3)3.
Answer:
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
