Question

cứu câu 28

Answer 1

Câu 28:

a, Ta có: \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PT: \(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)

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

\(\Rightarrow C\%_{HCl}=\dfrac{0,1.36,5}{150}.100\%=3,65\%\)

b, Theo PT: \(n_{BaCO_3}=n_{BaCl_2}=n_{CO_2}=0,05\left(mol\right)\)

⇒ \(n_{BaCl_2\left(bandau\right)}=\dfrac{15,05-0,05.197}{208}=0,025\left(mol\right)\)

⇒ n_BaCl2 = 0,05 + 0,025 = 0,075 (mol)

Ta có: m _{dd sau pư} = 15,05 + 150 - 0,05.44 = 162,85 (g)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,075.208}{162,85}.100\%\approx9,58\%\)