Answer:
0.4402 grams of Mg is needed to displace all of the copper (II) ions from the solution.
Explanation:
Moles of CuO = [tex]\frac{1.4583 g}{79.5 g/mol}=0.01834 mol[/tex]
1 mole of CuO has 1 mole of [tex]Cu^{2+}[/tex] ions, then 0.01834 molesof CuO will have 0.01834 moles of
[tex]Mg(s) + Cu^{2+}(aq)\rightarrow Cu(s) + Mg^{2+}(aq)[/tex]
According to reaction , 1 mole of copper (II) ion reacts with 1 mole of magnesium solid .
Then 0.01834 moles of copper (II) ion will react with :
[tex]\frac{1}{1}\times 0.01834 mol=0.01834 mol[/tex] of magnesium.
Mass of 0.01834 moles of magnesium :
0.01834 mol × 24 g/mol = 0.4402 g
0.4402 grams of Mg is needed to displace all of the copper (II) ions from the solution.
