\(Gọi:n_{Fepứ}=x\left(mol\right)\\ Fe+CuSO_4\rightarrow FeSO_4+Cu\\ m_{tăng}=m_{Cu}-m_{Fe}=64x-56x=1,2\\ \Rightarrow x=0,15\left(mol\right)\\ \Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\\ \Rightarrow ChọnD\)

D

\(n_{KOH}=0,04.2=0,08\left(mol\right)\)

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

______0,08------------------->0,04

=> m_Cu(OH)2 = 0,04.98 = 3,92(g)

=> A

\(n_{KOH}=2.0,04=0,08\left(mol\right)\)

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

______0,08------------------->0,04

=> m_Cu(OH)2 = 0,04.98 = 3,92(g)

=> A

\(m_{KNO_3}=\frac{40.45}{100}=18\left(g\right)\)

\(m_{KNO_3}=\frac{40.45}{100}=18\left(g\right)\)

help

Bài 1 chưa hết đề ở phần cuối: )

2

\(\Sigma m_{KNO_3\left(ban.đầu\right)}=\dfrac{61,9m}{\left(61,9+100\right)}=0,3823m\left(g\right)\)

\(\Sigma m_{KNO_3\left(sau\right)}=118,2+\dfrac{21,9\left(m-118,2\right)}{21,9+100}=0,1797m+96,9647\left(g\right)\)

\(\Rightarrow0,3823m=0,1797m+96,9647\\ \Rightarrow m=478,6\left(g\right)\)

\(Đặt:n_{Mg}=a\left(mol\right),n_{Zn}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+65b=15,4\\2a+2b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ m_{muối}=m_{MgCl_2}+m_{ZnCl_2}=95a+136b=36,7\left(g\right)\\ \Rightarrow A\)