The following balanced equations represent double replacement reactions that each result in the formation of precipitation. for each reaction, which ratio of reactants would result in the formation of the greatest amount of precipitate? cu(no3)2(aq) -> cu(oh)2(s)+ 2nano3 (aq) note: cu(oh)2 is a blue precipitate.
feso4 (aq) + 2naoh (aq) -> fe(oh)2 (s) + na2so4 (aq) note: fe(oh)2 is a dark green precipitate.

fe(no3)3 (aq) + 3naoh (aq) -> fe(oh)3 (s) + 3nano3 (aq) note: fe(oh)3 is a red-orange precipitate

1) Cu(NO₃)₂(aq) + 2NaOH (aq) -> cu(OH)₂(s)+ 2NaNO₃ (aq)

   reactants ratio = Cu(NO₃)2 : NaOH  = 1:2

2) FeSO₄ (aq) + 2NaOH (aq) ->Fe(OH)₂ (s) + Na₂SO₄ (aq)

  reactants ratio = FeSO₄  : NaOH  = 1:2

3) Fe(NO₃)₃ (aq) + 3NaOH (aq) -> Fe(OH)₃ (s) + 3NaNO₃ (aq)

  reactants ratio = Fe(NO₃)₃  : NaOH  = 1:3

Explanation:

For any reaction to occur completely, the ratio of the reactants should be the ratio of their coefficients is the balanced chemical equation.

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.

In our case, we get greatest amount of precipitate only when the reactants completely undergo chemical change and turn into products.

so, when reactants are taken in the ration of balanced equation, the reactants completely form products, where here one of the products is precipitate.

1) Cu(NO₃)₂(aq) + 2NaOH (aq) -> cu(OH)₂(s)+ 2NaNO₃ (aq)

   reactants ratio = Cu(NO₃)2(aq) : NaOH (aq) = 1:2

2) FeSO₄ (aq) + 2NaOH (aq) ->Fe(OH)₂ (s) + Na₂SO₄ (aq)

  reactants ratio = FeSO₄ (aq) : NaOH (aq) = 1:2

3) Fe(NO₃)₃ (aq) + 3NaOH (aq) -> Fe(OH)₃ (s) + 3NaNO₃ (aq)

  reactants ratio = Fe(NO₃)₃ (aq) : NaOH (aq) = 1:3  

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

2b2t

explanation:

2b2t