thôi bó tay 

rồi đề bắt tính chi :))

\(a,n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,2}{0,4}=0,5M\\ b,n_{H_2SO_4}=\dfrac{73,5}{98}=0,75\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,75}{0,5}=1,5M\\ n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,4}{0,25}=1,6M\\ n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\\ C_{M\left(Ba\left(OH\right)_2\right)}=\dfrac{0,2}{0,8}=0,25M\)

KMnO₄

KMnO₄

2Al+3H2SO4->Al2(SO4)3+3H2O

0,15------------------0,075--------0,225

n Al=0,15 mol

=>m Al2(SO4)3=0,075.342=25,65g

=>VH2=0,225.22,4=5,04l

Đề yêu cầu tính gì vậy bạn ?

\(n_{Ba}=\dfrac{109,6}{137}=0,8\left(mol\right)\\ m_{ddHCl}=200.1,1=220\left(g\right)\\ \rightarrow m_{HCl}=20\%.220=44\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{44}{36,5}=1,205\left(mol\right)\)

PTHH: Ba + 2HCl ---> BaCl₂ + H₂

LTL: \(0,6< \dfrac{1,205}{2}\rightarrow\) HCl dư

Theo pthh: \(n_{H_2}=n_{Ba}=n_{BaCl_2}=0,6\left(mol\right)\)

\(\rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)

\(m_{dd}=220+109,6=329,6\left(g\right)\)

\(m_{BaCl_2}=0,6.208=124,8\left(g\right)\)

Theo pthh: \(n_{HCl\left(pư\right)}=2n_{Ba}=2.0,6=1,2\left(mol\right)\)

=> \(\left\{{}\begin{matrix}C\%_{BaCl_2}=\dfrac{124,8}{329,6}=36,86\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,05.36,5}{329,6}=0,55\%\end{matrix}\right.\)

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

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

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