KMnO₄

KMnO₄

\(n_{Al}=\dfrac{8,1}{27}=0,3mol\)

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

0,3       0,45            0,15              0,45

a)\(m_{H_2SO_4bđ}=0,45\cdot98=44,1g\)

b)\(m_{ddH_2SO_4}=\dfrac{44,1}{10\%}\cdot100\%=441g\)

\(m_{H_2}=0,45\cdot2=0,9g\)

\(BTKL:m_{Al}+m_{ddH_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=8,1+441-0,9=448,2g\)

c)\(m_{Al_2\left(SO_4\right)_3}=0,15\cdot342=51,3g\)

\(C\%=\dfrac{51,3}{448,2}\cdot100\%=11,44\%\)

a) \(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,05-->0,1---->0,05-->0,05

=> V_H2= 0,05.22,4 = 1,12 (l)

b) \(\left\{{}\begin{matrix}C_{M\left(ZnCl_2\right)}=\dfrac{0,05}{0,2}=0,25M\\C_{M\left(HCl.dư\right)}=\dfrac{0,2-0,1}{0,2}=0,5M\end{matrix}\right.\)

\(n_K=\dfrac{9,75}{39}=0,25\left(mol\right)\)

\(n_{HCl}=\dfrac{300.7,3\%}{36,5}=0,6\left(mol\right)\)

PTHH: 2K + 2HCl --> 2KCl + H₂

Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,6}{2}\) => HCl dư

PTHH: 2K + 2HCl --> 2KCl + H₂

          0,25-->0,25-->0,25-->0,125

=> V_H2 = 0,125.22,4 = 2,8 (l)

m_KCl = 0,25.74,5 = 18,625 (g)

m_{dd sau pư} = 9,75 + 300 - 0,125.2 = 309,5 (g)

m_HCl(dư) = (0,6 - 0,25).36,5 = 12,775 (g)

\(\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{18,625}{309,5}.100\%=6,018\%\\C\%_{HCl}=\dfrac{12,775}{309,5}.100\%=4,128\%\end{matrix}\right.\)

bn tham khảo

TL: 

Tham khảo nhé: 

@@@@@@@@@@@@@@@@@@@@ 

@tuantuthan 

HT

\(n_{hhkhí}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

Gọi \(n_{SO_2}=a\left(mol\right)\left(0< a< 0,75\right)\)

\(\rightarrow n_{O_2\left(dư\right)}=0,75-b\left(mol\right)\)

Ta có: \(\dfrac{64a+32\left(0,75-a\right)}{0,75}=\dfrac{33,6}{1}=33,6\left(\dfrac{g}{mol}\right)\)

\(\rightarrow a=0,0375\left(mol\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,0375}{0,75}=5\%\\\%V_{O_2\left(dư\right)}=100\%-5\%=95\%\end{matrix}\right.\)

Fe+H2SO4-->FeSO4+H2

x---------------------x mol

Mg+H2SO4-->MgSO4+H2

y----------------------------y mol

nH2=\(\dfrac{8,96}{22,4}\)=0,4 mol

ta có hệ pt

56x+24y=16

x+y=0,4

==>x=0,2 mol, y=0,2 mol

=>%mFe=(\(\dfrac{0,2.56}{16}\).100=70%

=>%mMg=100-70=30%

=>m muoi=0,2.152+0,2.120=176,2g

Mình sẽ chon hoá trị cao nhất của P bạn nhé :) 

PTHH:

\(4P+3O_2\underrightarrow{t^o}2P_2O_3\\ P_2O_3+O_2\underrightarrow{t^o}P_2O_5\)

Giả sử mỗi ddbh có 100 g nước 

\(\rightarrow m_{KNO_3\left(\text{kết tinh}\right)}=61,3-21,9=39,4\left(g\right)\)

Mà theo đề bài: \(m_{KNO_3\left(\text{kết tinh}\right)}=118,2\left(g\right)\)

\(\rightarrow m_{H_2O\left(\text{thực tế}\right)}=\dfrac{118,2}{39,4}.100=300\left(g\right)\)

Gọi \(m_{KNO_3\left(ddbh:40^oC\right)}=a\left(g\right)\)

\(\rightarrow S_{KNO_3\left(ddbh:40^oC\right)}=\dfrac{a}{300}.100=61,3\left(g\right)\\ \rightarrow a=183,9\left(g\right)\\ \rightarrow m=300+183,9=483,9\left(g\right)\)

Giả sử mỗi ddbh có 100 g nước 

\(\rightarrow m_{KNO_3\left(\text{kết tinh}\right)}=61,3-21,9=39,4\left(g\right)\)

Mà theo đề bài: \(m_{KNO_3\left(\text{kết tinh}\right)}=118,2\left(g\right)\)

\(\rightarrow m_{H_2O\left(\text{thực tế}\right)}=\dfrac{118,2}{39,4}.100=300\left(g\right)\)

Gọi \(m_{KNO_3\left(ddbh:40^oC\right)}=a\left(g\right)\)

\(\rightarrow S_{KNO_3\left(ddbh:40^oC\right)}=\dfrac{a}{a+300}=61,3\left(g\right)\\ \rightarrow a=305\left(g\right)\\ \rightarrow m=a+300=305+300=605\left(g\right)\)

Nồng độ \(CaCl_2\) trong \(CaCl_2.6H_2O\) là:

\(C\%=\dfrac{M_{CaCl_2}}{M_{CaCl_2.6H_2O}}\cdot100\%=\dfrac{111}{219}\cdot100\%=50,68\%\)

Sơ đồ chéo:

\(CaCl_2.6H_2O\)    \(m_1\)    50,68                   40

                                          40

\(H_2O\)                 \(m_2\)    0                          10,68

\(\Rightarrow\dfrac{m_1}{m_2}=\dfrac{40}{10,68}\)  (*)

\(m_{dd}=V\cdot D=10\cdot1,395=13,95g=m_1+m_2\)

\(\Rightarrow m_2=13,95-m_1\) Thay vào (*) ta được:

\(\Rightarrow\dfrac{m_1}{13,95-m_1}=\dfrac{40}{10,68}\Rightarrow m_1=11,01g\)

\(\Rightarrow m_2=13,95-11,01=2,94g\Rightarrow n_{H_2O}=0,163mol\)

\(V_{H_2O}=0,163\cdot22,4=3,65l\)