Question

An 8.00 g sample of a copper ore was dissolved in an acid solution and then reacted with NO3 − from a KNO3 salt solution. 3 Cu + 8 NO− 3 + 8 H+ → 3 Cu2+ + 2 NO + 4 H2O + 6 NO− 3 It required 71.5 mL of a 0.155 M KNO3 solution to fully react the copper in the ore sample. What was the percent copper in the ore?

Answer 1

3.30%

Step-by-step explanation:

First lets calculate the moles of
`NO_3^(-)` (from the
`KNO_3` used in the reaction based on the volume and molarity of solution.

Moles
`KNO_3` = molarity * volume of
`KNO_3`

`n_(KNO_3) = (0.155 M) (0.0715 L)=0.0111 mol KNO_3`

Based on the stoichiometry, we can calculate the moles and mass of Cu.

`m_(Cu)=0.0111 mol NO_3^(-) *(3 mol Cu)/(8 mol NO_3^-) *(63.55 g Cu)/(mol Cu) = 0.265 g Cu`

Since we have that 8.00 g of ore contains 0.265 g of Cu, the mass percentage is:

Percentage of Cu in the ore = (0.265 g /8.00 g) x 100% = 3.30%