\(a.2KOH+CuSO_4\rightarrow K_2SO_4+Cu\left(OH\right)_2\\ b.n_{KOH}=\dfrac{70.11,2\%}{56}=0,14\left(mol\right) \\ n_{CuSO_4}=n_{Cu\left(OH\right)_2}=n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,07\left(mol\right)\\ V_{ddCuSO_4}=\dfrac{0,07}{1}=0,07\left(l\right)=70ml \\ m_{ddCuSO_4}=70.1,31=91,7\left(g\right)\\ m_{ddsaupu}=70+91,7-0,07.98=154,84\left(g\right)\\ C\%_{K_2SO_4}=\dfrac{0,07.174}{154,84}.100=7,87\%\)
a)b)
\(m_{ct\left(KOH\right)}=\dfrac{m_{dd}\cdot C\%}{100\%}=\dfrac{70\cdot11,2\%}{100\%}=7,84\left(g\right)\)
\(n_{ct\left(KOH\right)}=\dfrac{7,84}{39+16+1}=0,14\left(mol\right)\)
PTHH:
\(2KOH+CuSO_4->Cu\left(OH\right)_2+K_2SO_4\)
0,14 --> 0,07 --> 0,07 --> 0,07 (mol)
\(V_{CuSO_4}=\dfrac{n}{C_M}=\dfrac{0,07}{1}=0,07\left(lít\right)=70\left(ml\right)\)
\(m_{dd\left(CuSO_4\right)}=d.V=1,31\cdot70=91,7\left(g\right)\)
\(m\left(ddsau\right)=m_{KOH}+m_{CuSO_4}-m_{Cu\left(OH\right)_2}\\ =70+91,7-6,86=154,84\left(g\right)\)
Chất thu được sau phản ứng là \(K_2SO_4\)
\(C\%_{K_2SO_4}=\dfrac{m_{ct}}{mddsau}\cdot100\%=\dfrac{0,07\cdot174}{154,84}\cdot100\%\approx7,87\%\)
