Question

Đã trả lời bởi chuyên gia

Answer 1

a,

PTHH:

${BaCl}_{2} + H_{2} {SO}_{4} \to 2 HCl + {BaSO}_{4} ↓$

$n_{{BaCl}_{2}} = \frac{400.5, 2 %}{208} =$ 0,1 (mol)

$m_{dd H_{2} {SO}_{4}} = 100.1, 14 = 114 (g) n_{H_{2} {SO}_{4}} = \frac{114.20 %}{98} = 0, 2327 mol > n_{{BaCl}_{2}} = 0, 1 \Rightarrow {BaCl}_{2} hết, H_{2} {SO}_{4} dư Theo PTHH : \Rightarrow n_{{BaSO}_{4}} = n_{H_{2} {SO}_{4} pư} = n_{{BaCl}_{2}} = 0, 1 mol \Rightarrow m_{{BaSO}_{4}} = 0, 1.233 = 23, 3 (g) b, Theo PTHH : n_{HCl} = 2 . n_{{BaCl}_{2}} = 0, 2 mol m_{H_{2} {SO}_{4}} = 114.20 % = 22, 8 (g) \Rightarrow m_{H_{2} {SO}_{4} pu} = 0, 1.98 = 9, 8 (g) \Rightarrow m_{H_{2} {SO}_{4} dư} = 22, 8 - 9, 8 = 13 (g) \Rightarrow m_{dd spư} = m_{dd H_{2} {SO}_{4}} + m_{dd {BaCl}_{2}} - m_{{BaSO}_{4}} = 114 + 400 - 23, 3 = 490, 7 (g) \Rightarrow C %_{dd HCl} = \frac{0, 2.36, 5}{490, 7} . 100 % = 1, 59 % C %_{dd H_{2} {SO}_{4} dư} = \frac{13}{490, 7} . 100 % = 2, 65 %$

...Xem thêm

Answer 2

a,

PTHH:

${BaCl}_{2} + H_{2} {SO}_{4} \to 2 HCl + {BaSO}_{4} ↓$

$n_{{BaCl}_{2}} = \frac{400.5, 2 %}{208} =$ 0,1 (mol)

$m_{dd H_{2} {SO}_{4}} = 100.1, 14 = 114 (g) n_{H_{2} {SO}_{4}} = \frac{114.20 %}{98} = 0, 2327 mol > n_{{BaCl}_{2}} = 0, 1 \Rightarrow {BaCl}_{2} hết, H_{2} {SO}_{4} dư Theo PTHH : \Rightarrow n_{{BaSO}_{4}} = n_{H_{2} {SO}_{4} pư} = n_{{BaCl}_{2}} = 0, 1 mol \Rightarrow m_{{BaSO}_{4}} = 0, 1.233 = 23, 3 (g) b, Theo PTHH : n_{HCl} = 2 . n_{{BaCl}_{2}} = 0, 2 mol m_{H_{2} {SO}_{4}} = 114.20 % = 22, 8 (g) \Rightarrow m_{H_{2} {SO}_{4} pu} = 0, 1.98 = 9, 8 (g) \Rightarrow m_{H_{2} {SO}_{4} dư} = 22, 8 - 9, 8 = 13 (g) \Rightarrow m_{dd spư} = m_{dd H_{2} {SO}_{4}} + m_{dd {BaCl}_{2}} - m_{{BaSO}_{4}} = 114 + 400 - 23, 3 = 490, 7 (g) \Rightarrow C %_{dd HCl} = \frac{0, 2.36, 5}{490, 7} . 100 % = 1, 59 % C %_{dd H_{2} {SO}_{4} dư} = \frac{13}{490, 7} . 100 % = 2, 65 %$

1 câu trả lời 849

Câu hỏi hot cùng chủ đề

Bài viết mới nhất Lớp 9