a+b) PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{CO_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\\V_{CO_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)

c) Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=\dfrac{50\cdot40\%}{40}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) Tạo muối trung hòa, bazơ dư, tính theo CO₂

Bảo toàn Cacbon: \(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\) \(\Rightarrow m_{Na_2CO_3}=0,1\cdot106=10,6\left(g\right)\)

Tính khối lượng muối

Khối lượng NaOH có trong dung dịch :

m NaOH  = 40x50/100 = 20(gam) ứng với số mol là

m NaOH  = 20/40 = 0,5 mol

Số mol NaOH lớn gấp hớn 2 lần số mol  CO 2 , vậy muối thu được sẽ là  Na 2 CO 3

CO 2  + 2NaOH →  Na 2 CO 3  +  H 2 O

Theo phương trình hoá học, ta có :

n Na 2 CO 3 = n CO 2  = 0,1 mol

Khối lượng muối cacbonat thu được :  m Na 2 CO 3  = 106 x 0,1 = 10,6g

a, \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

b, \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\Rightarrow n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)

d, \(m_{NaOH}=550.10\%=55\left(g\right)\Rightarrow n_{NaOH}=\dfrac{55}{40}=1,375\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1,375}{0,1}=13,75>2\)

→ Pư tạo muối trung hòa Na₂CO₃.

PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

\(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

\(\left\{{}\begin{matrix}m_{Mg}=\dfrac{40.9}{100}=3,6\left(g\right)\\m_{Al}=9-3,6=5,4\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\end{matrix}\right.\)

PTHH:

Mg + 2HCl ---> MgCl₂ + H₂

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

2Al + 6HCl ---> 2AlCl₃ + 3H₂

0,2 ---------------------------> 0,3

\(\rightarrow V_{H_2}=\left(0,15+0,3\right).22,4=10,08\left(l\right)\)

\(n_{H_2O}=\dfrac{6,48}{18}=0,36\left(mol\right)\)

PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

Theo pthh: \(n_{O\left(oxit\right)}=n_{H_2\left(pư\right)}=n_{H_2O}=0,36\left(mol\right)\)

\(\rightarrow m_{oxit}=15,12+16.0,36=20,88\left(g\right)\)

\(n_{Fe}=\dfrac{15,12}{56}=0,27\left(mol\right)\)

CTHH: Fe_xO_y

=> x : y = 0,27 :  0,36 = 3 : 4

=> CTHH: Fe₃O₄ (oxit sắt từ)

mMg = 40%x9 = 3,6(g) =>nMg=3,6:24 = 0,15 (mol)
=> mAl = 9-3,6 = 5,4(g) => nAl = 5,4:27 = 0,2 (mol)
pthh : 2Al+6HCl -> 2AlCl3+3H2
         0,2                            0,3
        Mg+2HCl -> MgCl2 +H2
        0,15                          0,15 
=> nH2 = 0,15 + 0,3 = 0,45 (mol) 
=> VH2 = 0,45.22,4 = 10,08 (L) 
mH2 = 0,45 . 2 = 0,9 (mol) 
áp dụng BLBTKL  ta có : 
mH2 + moxit sắt = mFe + mH2O 
=> moxit sắt =  20,7 (g) 
 

\(1.\\ \left(a\right)n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{CaCO_3}=\dfrac{50}{100}=0,2\left(mol\right)\\ n_{Cl_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\\ n_{CO_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ n_{NaCl}=0,5.0,5=0,25\left(mol\right)\\ \left(b\right)m_{MgO}=0,15.40=6\left(g\right)\\ m_{Cl_2}=\dfrac{6,1975}{24,79}\cdot71=17,75\left(g\right)\\ \left(c\right)V_{hh}=\left(0,15.24,79\right)+\left(0,35.24,79\right)=12,395\left(l\right)\)

\(2,\\ \left(a\right)C_{\%NaCl}=\dfrac{40}{40+160}\cdot100=20\%\\ \left(b\right)C_{\%A}=\dfrac{0,5}{50}\cdot100=1\%\\ 3,\\ \left(a\right)C_{M\left(MgCl_2\right)}=\dfrac{0,5}{2,5}=0,2M\\ \left(b\right)C_{M\left(BaCl_2\right)}=\dfrac{0,2}{0,6:1,2}=0,4M\)

\(4,\\ \left(a\right)Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\\ n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{H_2}=n_{Fe}=0,2mol\\ V_{H_2}=0,2.24,79=4,958\left(g\right)\)

(b) mik nghĩ là tính m kim loại sau pư

\(CuO+H_2\xrightarrow[]{}Cu+H_2O\\ n_{Cu}=n_{H_2}=0,2mol\\ m_{Cu}=64.0,2=12,8\left(g\right)\)

\(n_{CaCO_3}=\dfrac{40}{100}=0,4\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,4\left(mol\right)\\ \Rightarrow n_{NaOH}=0,4.2=0,8\left(mol\right)\\ m_{NaOH}=0,8.40=32\left(g\right)\\ \Rightarrow m_{ddNaOH}=\dfrac{32.100}{25}=128\left(g\right)\\ \Rightarrow V_{ddNaOH}=\dfrac{m_{ddNaOH}}{D_{ddNaOH}}=\dfrac{128}{1,28}=100\left(ml\right)=0,1\left(l\right)\)

t6 là hạn chót đúng ko ? sát deadline quá mà