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: ?
2:3 And 3:1
To determine the ratio of reactants that would result in the formation of the greatest amount of precipitate for each reaction, you need to consider the stoichiometry of the balanced equations. The coefficients in the balanced equations represent the mole ratio between reactants and products.
Let's go through each reaction and analyze the stoichiometry:
1. Cu(NO3)2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaNO3(aq)
In this equation, the mole ratio between Cu(NO3)2 and Cu(OH)2 is 1:1, meaning 1 mole of Cu(NO3)2 reacts with 1 mole of Cu(OH)2 to form 1 mole of Cu(OH)2. Therefore, the ratio of reactants that results in the greatest amount of Cu(OH)2 precipitate is 1:1.
2. FeSO4(aq) + 2NaOH(aq) -> Fe(OH)2(s) + Na2SO4(aq)
In this equation, the mole ratio between FeSO4 and Fe(OH)2 is also 1:1, meaning 1 mole of FeSO4 reacts with 1 mole of Fe(OH)2 to form 1 mole of Fe(OH)2. Therefore, the ratio of reactants that results in the greatest amount of Fe(OH)2 precipitate is 1:1.
3. Fe(NO3)3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaNO3(aq)
In this equation, the mole ratio between Fe(NO3)3 and Fe(OH)3 is 1:1, meaning 1 mole of Fe(NO3)3 reacts with 1 mole of Fe(OH)3 to form 1 mole of Fe(OH)3. Therefore, the ratio of reactants that results in the greatest amount of Fe(OH)3 precipitate is 1:1.
In conclusion, for all the given reactions, the ratio of reactants that results in the formation of the greatest amount of precipitate is 1:1.
2JE
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
You answered correctly that 1 mol Cu(NO3)2 to 2 mols NaOH give you 1 mol ppt.
2. FeSO4(aq) + 2NaOH(aq) -> Fe(OH)2(s) + Na2SO4(aq) Note: Fe(Oh)2 is a dark green precipitate. Answer: ?
Why wouldn't this follow the same rule; i.e., 1 mol FeSO4 to 2 mols NaOH gives you 1 mol ppt?
3. Fe(NO3)3(aq) + 3NaOH (aq) -> Fe(OH)3 (s) + 3NaNO3(aq) Note: Fe(OH)3 is a red-orange precipitate. Answer: ?
Wouldn't this follow the same precedent?