a) BaCO₃ +2HCl \(\rightarrow\) BaCl₂ +H₂O + CO₂ \(\uparrow\)

b)\(n_{CO2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo ptpu : \(n_{BaCo3}=n_{Co2}=0,1\left(mol\right)\)

\(m_{BaCO3}=nM=0,1\left(137+12+3.16\right)=19,7\left(g\right)\)

\(m_{BaCl_2}=40,5-19,7=20,8\left(g\right)\)

c) \(\%m_{BaCO3}=\dfrac{19,7}{40,5}=48,6\%\)

\(\%m_{BaCl2}=100\%-48,6\%=51,4\%\)

d)Theo ptpu \(n_{BaCl2}=n_{Co2}=0,1\left(mol\right)\)

\(n_{HCl}=2n_{Co2}=0,2\left(mol\right)\)

\(m_{HCl}=0,2\left(1+35,5\right)=7,3\left(g\right)\)

\(m_{ddHCl}=\dfrac{mct.100\%}{C\%}=\dfrac{7,3.100}{10}=73\left(g\right)\)

\(m_{BaCl2}=20,8+0.1\left(137=2.35,5\right)=41,6\left(g\right)\)

\(m_{CO2}=0,1.\left(12+2.16\right)=4.4\left(g\right)\)

\(m_{ddBaCl2}=m_{hh}+m_{ddHCl}-m_{CO2}\)

\(=40,5+73-4,4=109,1\left(g\right)\)

\(C\%=\dfrac{mct}{mdd}.100\%=\dfrac{41,6}{109,1}100\%=38,1\%\)

đúng mà

10,35% VÀ 25,05G

CaCO₃ +2 HCl ➝ CaCl₂ + CO₂ + H₂O

a) Ta có

n\(_{CaCO3}=\frac{40}{100}=0,4\left(mol\right)\)

Theo pthh

n\(_{HCl}=2n_{CaCO3}=0,8\left(mol\right)\)

m\(_{HCl}=0,8.36,5=29,2\left(g\right)\)(Cái này là câu b r nha bạn)

m\(_{ddHCl}=\frac{29,2.100}{18,25}=160\left(g\right)\)

c) Ta có

Theo pthh

n\(_{CO2}=n_{CaCO3}=0,4\cdot\left(mol\right)\)=>mCO2 =0,4.44=17,6(g)

V\(_{CO2}=0,4.22,4=8.98\left(l\right)\)

mdd =160+29,2-17,6=171,6(g)

C%(CaCl2)=\(\frac{0,4.111}{171,6}.100\%=25,87\%\%\)

Chúc bạn học tốt

11.

n_Na = 0,4 mol

CH₃COOH + Na → CH₃COONa + H₂

⇒ m_CH3COOH = 0,4.60 = 24 (g)

⇒ V_H2 = 0,4.22,4 = 8,96 (l)

1. \(n_{NaCl}=x\left(mol\right)\)

\(PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

\(m_{ddspu}=200+120-22x=320-22x\left(g\right)\)

Theo đề bài ta có:

\(\frac{58,5x}{320-22x}.100\%=20\%\\ \Leftrightarrow.........................\\ \Leftrightarrow x=1,02\left(mol\right)\)

\(C\%_{Na_2CO_3}=\frac{\frac{1,02}{2}.106}{200}.100\%=27,03\left(\%\right)\)

\(C\%_{Na_2CO_3}=\frac{1,02.36,5}{120}.100\%=31,03\left(\%\right)\)

2. \(n_{NaCl}=x\left(mol\right)\)

\(PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

\(m_{ddspu}=307+365-22x=672-22x\left(g\right)\)

Theo đề bài ta có:

\(\frac{58,5x}{672-22x}.100\%=9\%\\ \Leftrightarrow.........................\\ \Leftrightarrow x=1\left(mol\right)\)

\(C\%_{Na_2CO_3}=\frac{\frac{1}{2}.106}{307}.100\%=17,26\left(\%\right)\)

\(C\%_{Na_2CO_3}=\frac{1.36,5}{365}.100\%=10\left(\%\right)\)

Bài1:
nNa2CO3 = x
Na2CO3 + 2HCl —> 2NaCl + CO2 + H2O
x…………….2x……………2x……..x
mdd sau phản ứng = mddNa2CO3 + mddHCl – mCO2 = 320 – 44x
C%NaCl = 58,5.2x/(320 – 44x) = 20%
—> x = 0,5087
C%Na2CO3 = 106x/200 = 26,96%
C%HCl = 36,5.2x/120 = 30,95%

Bài 2:
Gọi x là số mol của Na2CO3( chất tan)
Na2CO3 + 2HCl ---> 2NaCl + H2O + CO2
__x_______2x_______2x___________x
Ta có:
m NaCl = 117x (g)
m dd sau phản ứng = (307 + 365) - 44x ( mdd = m trươc p/ú - m khí )
Ta có: m ct / m dd = C / 100
=> 117x / (672 - 44x) = 9 \ 100
Giải ra x = 0.5(mol)
=> C% Na2CO3 = (0.5 x 106) / (672 - 44 x 0,5) x 100 = 8.15%
=> C% HCl = ( 2 x 0,5 x 36.5) / ( 672 - 44x0.5) x 100 =5.61%