Giúp e vsssssss ạhhhhhhhhh

Đáp án:

$a.$

$n_{KOH}=C_M \times V=C_M \times 300 \,\, mL  \times \dfrac{1 \,\, L}{1000 \,\, mL}=0.3C_M(mol)$

$n_{CuCl_2}=C_M \times V=2 \times 150 \,\, mL  \times \dfrac{1 \,\, L}{1000 \,\, mL}=0.3(mol)$

$2KOH+CuCl_2 \to Cu(OH)_2↓+2KCl$

$0.3C_M$`->`$0.15C_M$

`->`$0.15C_M=0.3$

`=>`$C_M=2(M)$

$b.$

$2KOH+CuCl_2 \to Cu(OH)_2↓+2KCl$

$0.6$      `->`     $0.3$    `->` $0.3$

`=>`$m_B=nM=0.3 \times 98=29.4(gam)$

$c.$

$Cu(OH)_2 \xrightarrow{t^o}CuO+H_2O$

$0.3$        `->`     $0.3$

`=>`$m_{rắn}=nM=0.3 \times 80=24.00(gam)$