Question

For the following unbalanced equation: Cu(s) + AgNO₃(aq) + Cu(NO₃)₂(aq) + Ag(s), if 1.90 moles of copper and 3.80 moles of silver nitrate are available to react, how many grams of silver are produced?

a) 107.87 g
b) 215.74 g
c) 431.48 g
d) 323.61 g

Answer 1

To find the number of moles of Cu²+ formed when 55.7 mol of H* are reacted, we can use the stoichiometry of the balanced equation. From the balanced equation, we can see that there is a 1:3 ratio between Cu(s) and Cu²+. Therefore, 20.81 mol of Cu²+ are formed.

Step-by-step explanation:

To find the number of moles of Cu²+ formed when 55.7 mol of H* are reacted, we can use the stoichiometry of the balanced equation.

The balanced equation is: 3Cu(s) + 2NO3¯(aq) + 8H*(aq) → 3Cu²+(aq) + 4H₂O(l) + 2NO(g)

From the balanced equation, we can see that there is a 1:3 ratio between Cu(s) and Cu²+. So, 3 moles of Cu²+ are formed for every 1 mole of Cu(s).

Therefore, to find the number of moles of Cu²+ formed, we can use the ratio:

(55.7 mol H*) x (3 mol Cu²+/8 mol H*) = 20.81 mol Cu²+