How do I find the mass of copper(II) hydroxide which is formed when excess copper(II) sulfate is added to 100mL of a 0.450 mol L^-1L−1 solution of sodium hydroxide?
1 Answer
Explanation:
An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide,
So, your sodium hydroxide solution contains
C = n/V implies n = C * VC=nV⇒n=C⋅V
n_(OH^(-)) = "0.450 M" * 100 * 10^(-3)"L" = "0.0450 moles OH"""^(-)nOH−=0.450 M⋅100⋅10−3L=0.0450 moles OH−
The net ionic equation for this double replacement reaction looks like this
"Cu"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-) -> "Cu"("OH")_text(2(s]) darrCu2+(aq]+2OH−(aq]→Cu(OH)2(s]⏐⏐↓
So,
This means that the reaction will produce
0.0450color(red)(cancel(color(black)("moles OH"""^(-)))) * ("1 mole Cu"("OH")_2)/(color(red)(2)color(red)(cancel(color(black)("moles OH"""^(-))))) = "0.0225 moles Cu"("OH")_2
To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass
0.0225color(red)(cancel(color(black)("moles OH"""^(-)))) * "97.56 g"/(1color(red)(cancel(color(black)("mole OH"""^(-))))) = "2.195 g Cu"("OH")_2
Rounded to three sig figs, the answer will be
m_(Cu(OH)_2) = color(green)("2.20 g")
