Bảo toàn khối lượng

\(m_{O_2}=m_{Oxit}-m_{KL}=48,84-34,44=14,4g\)

\(\rightarrow n_{O_2}=\frac{14,4}{32}=0,45mol\)

BTNT (O) \(n_{H_2O}=2n_{O_2}=0,9mol\)

\(n_{H_2}=\frac{4,032}{22,4}=0,18mol\)

BTNT (H)  \(n_{HCl}=2n_{H_2O}+2n_{H_2}=2,16mol\)

\(\rightarrow m_{HCl}=2,16.36,5=78,84g\)

\(m_{H_2}=0,18.2=0,36g\) và \(m_{H_2O}=0,9.18=16,2g\)

Bảo toàn khối lượng \(m_A+m_{HCl}=m_{\text{muối}}+m_{H_2O}+m_{H_2}\)

\(\rightarrow m_{\text{muối}}=48,84+78,84-0,36-16,2=111,12g\)

0,3mol chứ nhỉ?

a. \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\left(1\right)\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\left(2\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\left(3\right)\)

b. Theo phương trình \(n_{Al}=\frac{2}{3}n_{H_2}=0,2mol\) và \(n_{HCl\left(1\right)}=0,6mol\)

\(\rightarrow m_{FeO}+m_{Fe_2O_3}=35,8-0,2.27=30,4g\)

Đặt \(\hept{\begin{cases}n_{FeO}=x\\n_{Fe_2O_3}=y\end{cases}}\)

\(\rightarrow72x+160y=30,4\left(1\right)\)

Theo phương trình \(2x+6y=n_{HCl\left(2+3\right)}=1,6.1-0,6=1\left(2\right)\)

Từ (1) và (2) suy ra x = 0,2 và y = 0,1

\(\rightarrow m_{FeO}=0,2.72=14,4g\) và \(m_{Fe_2O_3}=0,1.160=16g\)

\(\rightarrow\%m_{FeO}=\frac{14,4}{35,8}.100\%\approx40,22\%\)

\(\rightarrow\%m_{Fe_2O_3}=\frac{16}{35,8}.100\%\approx44,69\%\)

c. Theo phương trình \(n_{AlCl_3}=0,2mol\) và \(n_{FeCl_2}=0,2mol\) và \(n_{FeCl_3}=0,2mol\)

\(\rightarrow m_{\text{muối}}=0,2.133,5+0,2.127+0,2.162,5=84,6g\)

BT1:

\(1,2FeS_2+\frac{11}{2}O_2\underrightarrow{t}Fe_2O_3+4SO_2\)

\(2,2SO_2+O_{_{ }2}\underrightarrow{t,V_2O_5}2SO_3\)

\(3,SO_3+H_2O\rightarrow H_2SO_4\)

\(4,H_2SO_4+Na_2SO_3\rightarrow Na_2SO_4+SO_2+H_2O\)

\(5,CaO+SO_2\underrightarrow{t}CaSO_3\)

\(6,CaSO_3\underrightarrow{t,xt}CaO+SO_2\)

\(7,CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(8,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

a) \(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)

PTHH : \(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\)    

Theo pthh : \(n_{H_2}=\frac{3}{2}n_{Al}=0,3\left(mol\right)\)

=> \(V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\)

b) Theo pthh :

\(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=0,1\left(mol\right)\)

 \(n_{H_2SO_4\left(pứ\right)}=\frac{3}{2}n_{Al}=0,3\left(mol\right)\) => \(V_{H_2SO_4}=\frac{\left(0,3+0,03\right)}{1}=0,33\left(l\right)\)

=> \(n_{H_2SO_4\left(dư\right)}=0,3\cdot0,1=0,03\left(mol\right)\)

=> \(\hept{\begin{cases}C_{M\left(Al_2\left(SO_4\right)_3\right)}=\frac{0,1}{0,33}\approx0,3\left(M\right)\\C_{M\left(H_2SO_4\right)}=\frac{0,03}{0,33}\approx0,09\left(M\right)\end{cases}}\)