Bạn đang học pha trộn dung dịch hả ?

Nếu không phải thì mình nghĩ đề là : 16g CuO -> 200g dd HCl

\(n_{HCl}=0,2.3=0,6\left(mol\right)\)

Gọi x số mol của FeO

Theo gt: \(n_{Fe_2O_3}=n_{FeO}=x\)

Pt: \(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\) (1)

x-------> x -------> 0,15

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\) (2)

x -------> 3x -------> 0,15

(1)(2) \(\Rightarrow x+3x=0,6\)

\(\Rightarrow x=0,15\left(mol\right)\)

\(C_{M_{FeSO_4}}=C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{0,15}{0,2}=0,75M\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Gọi \(\overline{M}\) là khối lượng hỗn hợp

ta có: \(M_A< \overline{M}< M_B\)

Pt: \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)

\(R+2H_2O\rightarrow R\left(OH\right)_2+H_2\)

\(\overline{M}=\dfrac{32}{0,3}=106,67\)

mà \(M_A< \overline{M}< M_B\)

\(\Rightarrow M_R< 106,67< 137\)

R là Kali

--------

Uầy không biết đúng không, nhìn sai sai /_/

\(n_{CO}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

\(n_{SO_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

Gọi R là kim loại cần tìm

cthc: \(R_xO_y;x,y\in Z^+\)

Pt: \(R_xO_y+yCO\rightarrow xR_{\dfrac{2y}{x}}+yCO_2\) (1)

0,8 ---------------- > 0,8

\(2R+2yH_2SO_{4\left(đ\right)}\underrightarrow{t^o}R_2\left(SO_4\right)_y+ySO_2+2yH_2O\)

1,8y <------------------------------ 0,9

\(m_R=46,6+0,8.28-0,8.44=33,8\left(g\right)\)

\(\Rightarrow m_R=1,8yM_R\)

\(\Leftrightarrow33,8=1,8y.M_R\)

\(\Rightarrow M_R=18,78y\)

Lập bảng ra thì tìm được M_R = 56 với y = 3

Từ 1 ta có: \(\dfrac{56x+48}{46,6}=\dfrac{3}{0,8}\)

\(\Rightarrow x=2\)

Vậy cthc: \(Fe_2O_3\)

\(n_{Al}=\dfrac{22,4}{27}=\dfrac{112}{135}\left(mol\right)\)

Pt: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(\dfrac{112}{135}\) ----> \(\dfrac{56}{45}\) ---------> \(\dfrac{56}{135}\) -------> \(\dfrac{56}{45}\)

\(V_{H_2}=\dfrac{56}{45}.22,4=27,87\left(l\right)\)

\(m_{Al_2\left(SO_4\right)_3}=\dfrac{56}{135}.342=141,87\left(g\right)\)

\(m_{H_2SO_4}=\dfrac{56}{45}.98\approx122\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{122.100}{300}=40,67\%\)

\(\Sigma_{m_{dd\left(spu\right)}}=22,4+300-\dfrac{56}{45}.2\approx320\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{141,87.100}{320}=46,21\%\)

\(\left(2x-5\right)^3-\left(x-2\right)^3=\left(x-3\right)^3\)

\(\Leftrightarrow\left(2x-5-x+2\right)\left(4x^2-20x+25+2x^2-9x+10+x^2-4x+4\right)=\left(x-3\right)^3\)

\(\Leftrightarrow\left(x-3\right)\left(7x^2-33x+39\right)-\left(x-3\right)^3=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x^2-33x+39-x^2+6x-9\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x^2-27x+30\right)=0\)

\(\Leftrightarrow\left(x-3\right)3\left(2x^2-9x+10\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x^2-4x-5x+10\right)=0\)

\(\Leftrightarrow\left(x-3\right)[2x\left(x-2\right)-5\left(x-2\right)]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\\2x-5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)

\(S=\left\{2;\dfrac{5}{2};3\right\}\)

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

Gọi x, y lần lượt là số mol của Al, Fe

Pt: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\) (1)

x ---> 3x/2 ---------> 0,1 ---------> 0,3

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\) (2)

y ---> y ----------> 0,1 -------> 0,1

(1)(2) \(\Rightarrow\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(V_{H_2\left(1\right)\left(2\right)}=\left(0,3+0,1\right).22,4=8,96\left(l\right)\)

\(m\)_Muối = \(m_{KL}+m_{axit}-m_{H_2}\)

\(=11+\left(0,3+0,1\right).98-\left(0,3+0,1\right).2=49,4\left(g\right)\)

\(2C_xH_y+\left(2x+\dfrac{1}{2}y\right)O_2\underrightarrow{t^o}2xCO_2+yH_2O\)

\(2C_2H_yO_z+\left(4-z+\dfrac{1}{2}y\right)O_2\underrightarrow{t^o}4CO_2+yH_2O\)

Nghe thú vị ghia ._.