Question

Answer 1

\(n_{CuSO_4}=1,5.0,1=0,15\left(mol\right)\)

\(n_{KOH}=1.0,2=0,2\left(mol\right)\)

PTHH: CuSO₄ + 2KOH --> Cu(OH)₂ + K₂SO₄

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,2}{2}\) => KOH hết, CuSO₄ dư

=> dd sau pư chứa K₂SO₄ và CuSO₄ dư

PTHH: CuSO₄ + 2KOH --> Cu(OH)₂ + K₂SO₄

               0,1<------0,2------>0,1------->0,1

=> \(\left\{{}\begin{matrix}C_{M\left(CuSO_{4\left(dư\right)}\right)}=\dfrac{0,15-0,1}{0,1+0,2}=\dfrac{1}{6}M\\C_{M\left(K_2SO_4\right)}=\dfrac{0,1}{0,1+0,2}=\dfrac{1}{3}M\end{matrix}\right.\)

m_kt = 0,1.98 = 9,8 (g)