When 20.0 ml of a 0.250 m (nh4)2s solution is added to 150.0 ml of a solution of cu(no3)2, a cus precipitate forms. the precipitate is then filtered from the solution, dried, and weighed. if the recovered cus is found to have a mass of 0.3491 g, what was the concentration of copper ions in the original cu(no3)2 solution?

The concentration fo copper in the Cu(NO₃)₂ solution is 0.0243 M.

Explanation:

Hi there!

Let´s write the balanced equation of the reaction:

(NH₄)₂S (aq) + Cu(NO₃)₂ (aq) → CuS (s) + 2NO₃⁻(aq) + 2NH₄⁺ (aq)

We obtain 0.3491 g of CuS. The molar mass of CuS is calculated adding the molar mass of Cu (63.546 g) plus the molar mass of S (32.065 g):

Molar mass CuS = 63.546 g + 32.065 g = 95.611 g

Then, if 1 mol CuS has a mass of 95,611 g, 0.3491 g of CuS will be the mass of (0.3491 g of CuS · 1 mol CuS / 95,611 g CuS) 3.651 × 10 ⁻³ mol CuS.

Looking at the chemical equation, we can notice that 1 mol of Cu(NO₃)₂ is needed to produce 1 mol CuS, then, the number of moles of Cu(NO₃)₂ in the solution will be equal to the number of moles of CuS obtained: 3.651 × 10 ⁻³ mol.

This number of moles is present in 0.1500 l, then, in 1 l there will be

(1.000 l · 3.651 × 10 ⁻³ mol / 0.1500 l) 0.0243 mol Cu(NO₃)₂

The concentration of Cu(NO₃)₂ is 0.0243 M. Since 1 mol of Cu(NO₃)₂ dissociates in 1 mol of Cu and 2 mol of NO₃, the concentration of Cu will be the same as the concentration of Cu(NO₃)₂ :  0.0243 M.

Let´s corroborate if this is possible. From the chemical equation, we know that 1 mol (NH₄)₂S reacts with 1 mol of Cu(NO₃)₂, so there must be at least 3.651 × 10 ⁻³ mol (NH₄)₂S to produce the amount of CuS obtained.

number of moles of (NH₄)₂S = 0.250 mol/l · 0.020 l = 5.00 × 10⁻³ mol

(NH₄)₂S is in excess (5.000 × 10⁻³ mol > 3.651 × 10 ⁻³ mol) , so that all the Cu(NO₃)₂ will react.

1 a

2 d

3 a

4 c

explanation:

i'm pretty sure its b i might be wrong though!