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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 :

KevinJoel

Answer:

3.30%

Explanation:

First lets calculate the moles of [tex]NO_3^{-}[/tex] (from the  [tex]KNO_3[/tex] used in the reaction based on the volume and molarity of solution.

Moles [tex]KNO_3[/tex] = molarity * volume of [tex]KNO_3[/tex]

[tex]n_{KNO_3} = (0.155 M) (0.0715 L)=0.0111 mol KNO_3[/tex]

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

[tex]m_{Cu}=0.0111 mol NO_3^{-} *\frac{3 mol Cu}{8 mol NO_3^-} *\frac{63.55 g Cu}{mol Cu} = 0.265 g Cu[/tex]

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%

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