Question

[Huỳnh Đăng Khoa]

Giúp mik bài 7,8 vs. Mik cám ơn

Answer 1

[𝓓𝓾𝔂 𝓐𝓷𝓱]

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

                0,1____0,3_____0,1_____0,15     (mol)

           \(2Al+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

               0,1______0,3__________0,05____0,15_____0,3    (mol)

           \(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)

              0,15_____0,3________0,15___0,15_____0,3   (mol)

Ta có: \(m_{Al}+m_{Cu}=0,1\cdot27+0,15\cdot64=12,3\left(g\right)\)           

Answer 2

[𝓓𝓾𝔂 𝓐𝓷𝓱]

Bài 7:

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)=n_{Mg}=n_{H_2}=n_{MgSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Ag}=15,6-0,2\cdot24=10,8\left(g\right)\\V_{H_2}=0,2\cdot22,4=2,24\left(l\right)\\C_{M_{MgSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)