a) \(m_{HCl}=200.10,95\%=21,9\left(g\right)\)

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

x_______2x________x____x(mol)

\(n_{HCl}=\dfrac{200.10,95\%}{36,5}=0,6\left(mol\right)\)

Dung dịch A phải có HCl dư mới có thể trung hòa được NaOH.

\(n_{NaOH}=0,05.2=0,1\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

y________y______y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}2x+y=0,6\\y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,25\\y=0,1\end{matrix}\right.\)

\(\Rightarrow a=m_{CaCO_3}=100x=100.0,25=25\left(g\right)\\ V=V_{CO_2\left(đktc\right)}=22,4x=22,4.0,25=5,6\left(l\right)\)

c) 

\(m_{ddA}=25+200-0,25.44=214\left(g\right)\\ C\%_{ddCaCl_2}=\dfrac{0,25.111}{214}.100\approx12,967\%\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,1.36,5}{214}.100\approx1,706\%\)

a) \(m_{HCl}=\dfrac{200.10,95}{100}=21,9\left(g\right)\)

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

b) \(n_{CaCO_3}=\dfrac{a}{100}=0,01a\left(g\right)\)

\(n_{NaOH}=0,05.2=0,1\left(mol\right)\)

PTHH: CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

______0,01a---->0,02a---->0,01a->0,01a___________(mol)

NaOH + HCl --> NaCl + H₂O

_0,1----->0,1___________________________________(mol)

=> 0,02a = 0,6 - 0,1

=> a = 25 (g)

c) \(V_{CO_2}=0,01.25.22,4=5,6\left(l\right)\)

d) \(\left\{{}\begin{matrix}C\%\left(CaCl_2\right)=\dfrac{0,25.111}{25+200-0,25.44}.100\%=12,97\%\\C\%\left(HCl_{dư}\right)=\dfrac{0,1.36,5}{25+200-0,25.44}.100\%=1,705\%\end{matrix}\right.\)

a. PTHH:

\(BaO+2HCl--->BaCl_2+H_2O\left(1\right)\)

\(BaCO_3+2HCl--->BaCl_2+CO_2\uparrow+H_2O\left(2\right)\)

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

Theo PT₍₂₎: \(n_{BaCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{BaCO_3}=0,2.197=39,4\left(g\right)\)

\(\Rightarrow\%_{m_{BaCO_3}}=\dfrac{39,4}{54,7}.100\%=72,03\%\)

\(\%_{m_{BaO}}=100\%-72,03\%=27,97\%\)

b. Ta có: \(m_{BaO}=54,7-39,4=15,3\left(g\right)\)

\(\Rightarrow n_{BaO}=\dfrac{15,3}{153}=0,1\left(mol\right)\)

\(\Rightarrow n_A=0,1+0,2=0,3\left(mol\right)\)

Theo PT_(1,2): \(n_{HCl}=2.n_A=2.0,3=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{21,9}{m_{dd_{HCl}}}.100\%=20\%\)

\(\Rightarrow m_{dd_{HCl}}=109,5\left(g\right)\)

\(n_{CO2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a) Pt : \(BaO+2HCl\rightarrow BaCl_2+H_2O|\)

             1            2               1           1

            0,1        0,2

            \(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O|\)

                  1             2             1            1          1

                 0,2         0,4                         0,2

\(n_{BaCO3}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{BaCO3}=0,2.197=39,4\left(g\right)\)

\(m_{BaO}=54,7-39,4=15,3\left(g\right)\)

⁰/₀BaO = \(\dfrac{15,3.100}{54,7}=27,97\)⁰/₀

⁰/₀BaCO₃ = \(\dfrac{39,4.100}{54,7}=72,03\)⁰/₀

b) Có : \(m_{BaO}=15,3\left(g\right)\)

\(n_{BaO}=\dfrac{15,3}{153}=0,1\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,2+0,4=0,6\left(mol\right)\)

\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{21,9.100}{20}=109,5\left(g\right)\)

 Chúc bạn học tốt

             nCO2= 4,48 / 22.4= 0,2 ( mol)

a, BaO + 2HCl -> BaCl2 + H2O      (1)

BaCO3 + 2HCl-> BaCl2 + H2O + CO2   (2)

0,2                                                  <- 0,2       (mol)

Theo pt (2) ta có nBaCO3= 0,2 => mBaCO3= 0,2 x 197= 39,4 (g)

=> mBaO= 54,7- 39,4= 15.3 (g)

->% BaCO3= ( 39,4 : 54,7)x 100%= 72%

->% BaO= 100%-72% = 28%

a)PTHH:     AgNO₃  + HCl → AgCl↓ + HNO₃ 

nHCl = 0,2.0,5 = 0,1 mol 

=> nAgCl = 0,1 mol  = nAgNO₃ = 0,1 mol  = nHCl phản ứng 

<=> mAgCl = 0,1.143,5 = 14,35 gam

mAgNO₃ = 0,1.170  = 17 gam

=> mdd AgNO₃  = \(\dfrac{17}{6,8\%}\)= 250 gam

b)    X +    2HCl --> XCl₂   + H₂ 

1,2 gam X tác dụng vừa đủ với 0,1 mol HCl

=> Số mol của 1,2 gam X = 0,05 mol

<=> M_x = \(\dfrac{1,2}{0,05}\)= 24  (g/mol)   =>   X là magie ( Mg )

Câu 1:

\(\text{a) }pthh:CaCO3+2HCl\rightarrow CaCl_2+CO_2+H_2O\left(1\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\left(2\right)\)

b) \(n_{NaOH}=C_M\cdot V=0,05\cdot2=0,1\left(mol\right)\)

Theo \(pthh\left(2\right):n_{HCl\left(2\right)}=n_{NaOH}=0,1\left(mol\right)\)

\(m_{HCl}=\dfrac{m_{d^2HCl}\cdot C\%}{100}=\dfrac{200\cdot10,95}{100}=21,9\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{m}{M}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\ \Rightarrow n_{HCl\left(1\right)}=0,6-0,1=0,5\left(mol\right)\)

Theo \(pthh\left(1\right):n_{CaCO_3}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=n\cdot M=0,25\cdot100=25\left(g\right)\)

c) Theo \(pthh\left(1\right):n_{CO_2}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)\)

\(\Rightarrow V_{CO_2}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\)

d) \(m_{CO_2}=n\cdot M=0,25\cdot44=11\left(g\right)\)

\(m_{HCl\left(dư\right)}=n\cdot M=0,1\cdot36,5=3,65\left(g\right)\)

\(m_{d^2A}=\left(m_{CaCO_3}+m_{d^2HCl}\right)-m_{CO_2}\\ =\left(25+200\right)-11=214\left(g\right)\)

\(\Rightarrow C\%\left(HCl_{dư}\right)=\dfrac{3,65\cdot100}{214}=1,71\%\)

a) \(m_{NaOH}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{200\cdot15}{100}=30\left(g\right)\)

\(m_{d^2A\text{ sau khi pha thêm }100\left(g\right)nước}=100+200=300\left(g\right)\)

\(\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{30\cdot100}{300}=10\%\)

b) \(m_{NaOH\text{ sau khi cho thêm }5\left(g\right)NaOH}=30+5=35\left(g\right)\)

\(m_{d^2B}=5+200=205\left(g\right)\)

\(\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{35\cdot100}{205}=17,07\%\)

c) \(C\%\left(NaOH\right)=\dfrac{30\cdot100}{150}=20\%\)

200ml = 0,2l

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

        1         2              1           1

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

a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

              1            1              1            1

            0,6          0,6

\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)

\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)

 Chúc bạn học tốt

a) PTHH : H₂SO₄ + 2KOH ----> K₂SO₄ + 2H₂O

b) n_H2SO4 = 0,2 . 1 = 0,2(mol)

Theo pthh : n_KOH = 2n_h2so4 = 0,4 (mol)

--> m_KOH = 0,4.56 = 22,4 (g)

---> m_(ddKOH)  = 22,4 : 6 . 100 = 373,33 (g)

---> V_(ddKOH) = 373,33 : 1,048 = 356,23 (ml)

p/s: check lại hộ mình phát =)))