\(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\ PTHH:CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\\ \Rightarrow n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\\ \Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\)

\(n C a C O 3 = 10 100 = 0 , 1 ( m o l ) P T H H : C a C O 3 + 2 H C l → C a C l 2 + C O 2 ↑ + H 2 O ⇒ n H C l = 2 n C a C O 3 = 0 , 2 ( m o l ) ⇒ m H C l = 0 , 2 ⋅ 36 , 5 = 7 , 3 ( g )\)

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

\(PTHH:CaCO_3+2HCl--->CaCl_2+CO_2+H_2O\)

Theo PT: \(n_{HCl}=2.n_{CaCO_3}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)

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

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)

Câu 1 :

$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2$
$Fe + 2HCl \to FeCl_2 + H_2$
$FeCl_2 + 2KOH \to Fe(OH)_2 + 2KCl$
Câu 2 : 

$a) Zn + 2HCl \to ZnCl_2 + H_2$

b) Theo PTHH : $n_{H_2} = n_{Zn} = \dfrac{32,5}{65} = 0,5(mol)$
$V_{H_2} = 0,5.22,4 = 11,2(lít)$

c) $n_{ZnCl_2} = 0,5(mol) \Rightarrow m_{ZnCl_2} = 136.0,5 = 68(gam)$

d) $n_{HCl} = 2n_{Zn} = 1(mol) \Rightarrow C_{M_{HCl}} = \dfrac{1}{0,4} = 2,5M$

PTHH: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a) Theo PT: \(n_{CaCO_3}pư=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}pư=0,1\times100=10\left(g\right)\)

b) Theo PT: \(n_{HCl}pư=2n_{CO_2}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{HCl}pư=0,2\times36,5=7,3\left(g\right)\)

c) Theo PT: \(n_{CaCl_2}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCl_2}=0,1\times111=11,1\left(g\right)\)

a. Số mol \(CaCO_3\) tham gia phản ứng là:
\(n_{CaCO_3}=\dfrac{100}{100}=1\left(mol\right)\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
Theo phương trình thì ta có:
\(n_{HCl}=\dfrac{1.2}{1}=2\left(mol\right)\)\(\Rightarrow m_{HCl}=n.M=2.36,5=73\left(g\right)\)
\(n_{CaCl_2}=1\left(mol\right)\Rightarrow m_{CaCl_2}=111\left(g\right)\)
b. \(\Rightarrow\) \(n_{CO_2}=1\left(mol\right)\Rightarrow V_{CO_2}=1.22,4=22,4\left(l\right)\)

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

\(n_{CaCO_3}=\dfrac{100}{100}=1\left(mol\right)\)

a) Theo PT: \(n_{HCl}=2n_{CaCO_3}=2\times1=2\left(mol\right)\)

\(\Rightarrow m_{HCl}=2\times36,5=73\left(g\right)\)

Theo PT: \(n_{CaCl_2}=n_{CaCO_3}=1\left(mol\right)\)

\(\Rightarrow m_{CaCl_2}=1\times111=111\left(g\right)\)

b) Theo PT: \(n_{CO_2}=n_{CaCO_3}=1\left(mol\right)\)

\(\Rightarrow V_{CO_2}=1\times22,4=22,4\left(l\right)\)