Thiếu số gam oxit rồi bạn ơi

PT: \(R_2CO_3+2HCl\rightarrow2RCl+H_2O+CO_2\)

Ta có: \(n_{R_2CO_3}=\dfrac{21,2}{2M_R+60}\left(mol\right)\)

\(n_{RCl}=\dfrac{23,4}{M_R+35,5}\left(mol\right)\)

Theo PT: \(n_{RCl}=2n_{R_2CO_3}\)

\(\Rightarrow\dfrac{23,4}{M_R+35,5}=\dfrac{42,4}{2M_R+60}\)

\(\Rightarrow M_R=23\left(g/mol\right)\)

Vậy: R là Na.

Ta có: \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Na_2CO_3}=0,4\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

Bạn tham khảo nhé!

\(2A+2aHCl\rightarrow2ACl_a+aH_2\)

\(TheoPTHH:n_{H_2}=\dfrac{an_A}{2}=0,5an_A=0,5.a.\dfrac{1,17}{A}=\dfrac{V}{22,4}=0,015\)

\(\Rightarrow\dfrac{a}{A}=\dfrac{1}{39}\)

\(\Rightarrow\) A là Kali ( K ).

\(\Rightarrow n_{HCl}=2n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,03}{1,2}=0,025\left(l\right)=25\left(ml\right)\)

PT: \(R_2O_3+6HCl\rightarrow2RCl_3+3H_2O\)

Ta có: \(m_{HCl}=\dfrac{109,5.20}{100}=21,9\left(g\right)\)

\(\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Theo PT: \(n_{R_2O_3}=\dfrac{1}{6}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow M_{R_2O_3}=\dfrac{16}{0,1}=160\left(g/mol\right)\)

\(\Rightarrow2M_R+16.3=160\)

\(\Rightarrow M_R=56\left(g/mol\right)\)

Vậy: Đó là Fe.

Bạn tham khảo nhé!

Gọi R là kim loại cần tìm

nH2 = 0,8 mol

                R + 2HCl -> RCl2 + H2

(mol)      0,8 <-                        0,8

Ta có: R. 0,8 = 19,2 => R = 24 (Mg)

Gọi n là hóa trị của A

\(n_{H_2} = \dfrac{0,336}{22,4}=0,015(mol)\\ 2A + 2nHCl \to 2ACl_n + nH_2\\ n_A = \dfrac{2}{n}n_{H_2} = \dfrac{0,03}{n}(mol)\\ M_A = \dfrac{1,17}{\dfrac{0,03}{n}}=39n\)

Với n = 1 thì A = 39(Kali)

Vậy A là Kali

\(n_{HCl} = 2n_{H_2} = 0,03(mol)\\ m_{HCl} = 0,03.36,5 = 1,095(gam)\)

a) Ta có: \(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

Gọi a và b lần lượt là số mol của Al và Zn

Bảo toàn mol e: \(3a+2b=1,4\)

                   Mà \(27a+65b=31,4\)

\(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{31,4}\cdot100\%\approx17,2\%\\\%m_{Zn}=82,8\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=1,4mol\)

\(\Rightarrow V_{HCl}=\dfrac{1,4}{2}=0,7\left(l\right)=700\left(ml\right)\)

Đặt : 

nAl = a mol 

nZn = b mol 

mB = 27a + 65b = 31.4 (g) (1) 

2Al + 6HCl => 2AlCl3 + 3H2 

a___________________1.5a

Zn + 2HCl => ZnCl2 + H2 

b__________________b

nH2 = 1.5a + b = 15.68/22.4 = 0.7 (mol) (2) 

(1) , (2) : 

a = 0.2

b = 0.4 

%Al = 5.4/31.4 * 100% = 17.19%

%Zn = 100 - 17.19 = 82.81%

nHCl = 2nH2 = 0.7*2 = 1.4 (mol) 

Vdd HCl = 1.4 / 2 = 0.7 (l) 

PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\)  (1)

            \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)  (2)

a) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CaO}=0,1mol\\n_{CaCO_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,1\cdot56=5,6\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{hh}=20+5,6=25,6\left(g\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=0,2mol\\n_{HCl\left(2\right)}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=0,6mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

PT: \(CaO+2HCl\rightarrow CaCl_2+H_2O\) (1)

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

a, Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\)

Theo PT (2): \(n_{CaCl_2}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow n_{CaCl_2\left(1\right)}=0,3-0,2=0,1\left(mol\right)\)

Theo PT (1): \(n_{CaO}=n_{CaCl_2}=0,1\left(mol\right)\)

Theo PT (2): \(n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_A=m_{CaO}+m_{CaCO_3}=0,1.56+0,2.100=25,6\left(g\right)\)

b, Theo PT (1) + (2): \(\Sigma n_{HCl}=2n_{CaO}+2n_{CaCO_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,6}{0,3}=2M\)

Bạn tham khảo nhé!

PTHH: MO+2HCl---->MCl2+H2O

Ta có

n\(_{MO}=\frac{15,3}{M+16}\left(mol\right)\)

n\(_{MCl2}=\frac{20,8}{M+71}\)(mol)

Theo pthh

n M=n MCl2

-->\(\frac{15,3}{M+16}\) \(=\frac{20,8}{M+71}\)

-->15,3M+1086,3=20,8M+332,8

-->5,5M=753,5

-->M=137(Ba)

Vậy Oxxi kim loại đó là BaO

n BaO=15,3/153=0,1(mol)

Theo pthh

n HCl=2n BaO=0,2(mol)

m HCl=0,2.36,5=7,1(g)

m dd HCl=7,1.100/18,25=38,9(g)

\(MO+2HCl\rightarrow MCl_2+H_2O\)

0,1_____0,2____________________

\(n_{MCl2}=\frac{20,8}{M+71}\)

\(n_{MO}=\frac{15,3}{M+16}\)

Ta có nMO=nMCl2

\(\Leftrightarrow\frac{15,3}{M+16}=\frac{20,8}{M+71}\)

\(\Leftrightarrow M=137\left(Ba\right)\)

\(n_{Ba}=\frac{15,3}{137+16}=0,1\)

\(m_{dd_{HCl}}=\frac{0,2.36,5}{18,25\%}=40\left(g\right)\)