Respuesta :
In this case, we have NaOH as the limiting agent. Therefore the reaction will be according to the number of moles of NaOH present. We have the following amount of NaOH:
The molecular weight (MW) of NaOH: 40.0 g/mol
[tex]\begin{gathered} \text{NaOH mol: }\frac{\text{Mass of NaOH}}{\text{MWof NaOH}} \\ \text{NaOH mol: }\frac{\text{2}80.0\text{ g}}{\text{4}0.0\text{ g/mol}}=\text{ 7 mol of NaOH} \end{gathered}[/tex]We have 7 mol of NaOH, so we will have 3.5 mol of Cu(OH)2. Because the ratio NaOH: Cu(OH)2 is 2:1. Now, we have to determine the grams of Cu(OH)2 using the molecular weight:
[tex]\begin{gathered} \text{Mass of Cu(OH)2 }=\text{ MW Cu(OH)2 }\times\text{ Mol of Cu(OH)2} \\ \text{Mass of Cu(OH)2 }=\text{ 97.6 g/mol }\times\text{ 3}.5\text{ mol of Cu(OH)2} \\ \text{Mass of Cu(OH)2 }=\text{ 341.6 g of Cu(OH)2} \end{gathered}[/tex]So, we will have 341.6 g of Cu(OH)2 precipitated after the reaction
