Question

5. The following balanced equations represent double replacement reactions that each result in the formation of a precipitate. For each reaction, which ratio of reactants would result in the formation of the greatest amount of precipitate? I found the first one
1. Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaNO3(aq) Note: Cu(OH)2 is a blue precipitate. Answer 1:2
2. FeSO4(aq) + 2NaOH(aq) -> Fe(OH)2(s) + Na2SO4(aq) Note: Fe(Oh)2 is a dark green precipitate. Answer: ?
3. Fe(NO3)3(aq) + 3NaOH (aq) -> Fe(OH)3 (s) + 3NaNO3(aq) Note: Fe(OH)3 is a red-orange precipitate. Answer: ?

Answer 1

2:3 And 3:1

Answer 2

I don't understand your problem here. You did the first one correctly. How are the others any different?
1. Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaNO3(aq) Note: Cu(OH)2 is a blue precipitate. Answer 1:2
<b>You answered correctly that 1 mol Cu(NO3)2 to 2 mols NaOH give you 1 mol ppt.</b>
2. FeSO4(aq) + 2NaOH(aq) -> Fe(OH)2(s) + Na2SO4(aq) Note: Fe(Oh)2 is a dark green precipitate. Answer: ?
<b>Why wouldn't this follow the same rule; i.e., 1 mol FeSO4 to 2 mols NaOH gives you 1 mol ppt?</b>
3. Fe(NO3)3(aq) + 3NaOH (aq) -> Fe(OH)3 (s) + 3NaNO3(aq) Note: Fe(OH)3 is a red-orange precipitate. Answer: ?
<b> Wouldn't this follow the same precedent?</b>

Answer 3

2JE