Question

Giúp mình với

Answer 1

1)

$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$CO_2 + 2NaOH \to Na_2CO_3 + H_2O$

2) $n_{CaCO_3} = \dfrac{35}{100} = 0,35(mol)$

$n_{HCl} = 2n_{CaCO_3} = 0,7(mol)$
$m_{HCl} = 0,7.36,5 = 25,55(gam)$
$m_{dd\ HCl} = \dfrac{25,55}{25\%} = 102,2(gam)$

3)

$n_{CO_2} = n_{CaCO_3} = 0,35(mol)$
$n_{NaOH} = 2n_{CO_2} = 0,7(mol)$
$\Rightarrow V_{dd\ NaOH} = \dfrac{0,7}{0,5} = 1,4(lít) = 1400(ml)$

Answer 2

1)

\(n_{CaCO_3}=\dfrac{35}{100}=0,35\left(mol\right)\)

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

Mol:      0,35         0,7                     0,35

PTHH: CO₂ + 2NaOH → Na₂CO₃ + H₂O

Mol:     0,35      0,7

2) \(m_{ddHCl}=\dfrac{0,7.36,5.100}{25}=102,2\left(g\right)\)

3) \(V_{ddNaOH}=\dfrac{0,7}{0,5}=1,4\left(l\right)\)

Answer 3

Ta có: \(n_{CaCO_3}=\dfrac{35}{100}=0,35\left(mol\right)\)

1. PTHH:

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O (1)

CO₂ + 2NaOH ---> Na₂CO₃ + H₂O (2)

2. Theo PT₍₁₎: \(n_{HCl}=2.n_{CaCO_3}=2.0,35=0,7\left(mol\right)\)

=> \(m_{HCl}=0,7.36,5=25,55\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{25,55}{m_{dd_{HCl}}}.100\%=25\%\)

=> \(m_{dd_{HCl}}=102,2\left(g\right)\)

3. Theo PT₍₁₎; \(n_{CO_2}=n_{CaCO_3}=0,35\left(mol\right)\)

Theo PT₍₂₎: \(n_{NaOH}=2.n_{CO_2}=2.0,35=0,7\left(mol\right)\)

=> \(V_{dd_{NaOH}}=\dfrac{0,7}{0,5}=1,4\left(lít\right)=1400\left(ml\right)\)