n_CO2 = 0.1 (mol) 

Ba(OH)2 + CO2 => BaCO3 + H2O 

0.1________0.1_____0.1

CM Ba(OH)2 = 0.1/0.2 = 0.5 M 

mBaCO3 = 0.1*197 = 19.7 (g) 

Chúc học tốt <3 

Giúp mình vớiii

nH2 = 2.24/22.4 = 0.1 (mol) 

Fe + 2HCl => FeCl2 + H2 

0.1...............................0.1 

mFe = 0.1 * 56 = 5.6 (g) 

n_H = \(\dfrac{2,24}{22,4}\)= 0,1 (mol)

          Fe + 2HCl \(\rightarrow\)FeCl₂ + H₂

(mol)  0,1\(\leftarrow\)--------------------0,1

m_Fe = 0,1 . 56 = 5,6 (g)

Vậy x = 5,6

nH2 = 2.24/22.4 = 0.1 (mol) 

Fe + 2HCl => FeCl2 + H2 

0.1...............................0.1 

mFe = 0.1 * 56 = 5.6 (g) 

a)

\(Zn + 2HCl \to ZnCl_2 + H_2\)

b)

\(n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\)

Theo PTHH : \(n_{HCl} = 2n_{H_2} = 0,1.2 = 0,2(mol)\)

c)

Ta có :

 \(n_{ZnCl_2} = n_{H_2} = 0,1(mol)\\ \Rightarrow m_{ZnCl_2} = 0,1.136 = 13.6(gam)\)

Ta có: \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

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

               0,4 <---  0,6    ----------->      0,2    -->   0,6

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\\V_{H_2}=0,6.22,4=13,44\left(lít\right)\end{matrix}\right.\)

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

PTHH: Zn + 2HCl ---> FeCl₂ + H₂

          0,3<---------------0,3<----0,3

=> \(\left\{{}\begin{matrix}m=0,3.65=19,5\left(g\right)\\m_{muối}=0,3.136=40,8\left(g\right)\\V_{ddHCl}:thiếu.C_M\end{matrix}\right.\)

\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O
LTL: \(0,2>\dfrac{0,3}{3}\) => Fe₂O₃ dư

Theo pthh: n_{Fe2O3 (pư)} = \(\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

                  n_Fe = \(\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)

=> m_{chất rắn} = 0,1.160 + 0,2.56 = 27,2 (g)

Zn+2HCl->ZnCl2+H2

0,3---0,6----0,3----0,3

n H2=0,3 mol

=>m Zn=0,3.65=19,5g

=>m muối=0,3.136=40,8g

c) thiếu đề

d)

Fe2O3+3H2-to>2Fe+3H2O

               0,3------0,2

n Fe2O3=0,2 mol

=>Fe2O3 dư

=>m cr=0,2.56+0,1.160=27,2g

Theo ĐLBTKL: m_Zn + m_HCl = m_ZnCl2 + m_H2

=> m_HCl = 27,2 + 0,4 - 13 = 14,6 (g)

C

Theo định luật bảo toàn khối lượng,

         => 13 + HCl  = 27,1 + 0,4

         => HCl = 14,6

A

1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2

nH2(1)=3,36/22,4=0.15(mol)

=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)

2.Mg + 2HCl ==> MgCl2 + H2

3.2Al + 6HCl ==> 2AlCl3 + 3H2

4.Fe + 2HCl ==> FeCl2 + H2

=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)

=> nH2(3)=0.1*3/2=0.15(mol)

MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl

AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl

FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl

a) \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b) \(n_{Al}=\dfrac{9,72}{27}=0,36mol\)

Theo phương trình: \(n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}.0,36=0,27mol\)

\(\Rightarrow m_{O_2}=0,27.32=8,64g\)

c) Theo phương trình:  n_{\(Al_2O_3\)}  \(=\dfrac{1}{2}n_{Al}=0,18mol\)

\(\Rightarrow m_{Al_2O_3}=0,18.102=18,36g\)