Answer:
The answer to your question is:
a) Cu⁺² + 2NO₃⁻¹ + Na⁺¹ + 2OH⁻¹ ⇒ Cu⁺² + 2OH⁻¹ + Na⁺¹ + 2NO₃⁻¹
b) 2.45 g of Cu(OH)₂
Explanation:
Data
Copper (II) nitrate + Sodium hydroxide produces copper (II) hydroxide
(a) Write the net ionic equation for the reaction.
Cu(NO₃)₂ + 2NaOH ⇒ Cu(OH)₂ + 2NaNO₃
Cu⁺² + 2NO₃⁻¹ + Na⁺¹ + 2OH⁻¹ ⇒ Cu⁺² + 2OH⁻¹ + Na⁺¹ + 2NO₃⁻¹
(b) Calculate the maximum mass of copper(II) hydroxide that can be formed when 2.00 g of sodium hydroxide is added to 80.0 mL of 0.500 m Cu(NO3)2(aq).
Cu(NO₃)₂ + 2NaOH ⇒ Cu(OH)₂ + 2NaNO₃
MW NaOH = 23 + 16 + 1
= 40 g
40g of NaOH --------------- 1 mol
2 g ------------- x
x = (2 x 1) / 40
x = 0.05 mol
Molarity = moles / volume
moles = Molarity x volume
moles = 0.5 x 0.08
= 0.04 mol
The proportion is 1:2, 1 mol of Cu(NO₃)₂ to 2 mol of NaOH
Then the limiting reactant is NaOH because the proportion is 0.04 to 0.05 is not 1:2.
2 moles of NaOH ----------- 1 mol Cu(OH)₂
0.05 mol -------------------- x
x = (0.05 x 1) / 2
x = 0.025 mol of Cu(OH)₂
MW of Cu(OH)₂ = 64 + 32 + 2 = 98 g
98 g of Cu(OH)₂ --------------- 1 mol
x --------------- 0.025 mol
x = (0.025 x 98) / 1
x = 2.45 g
