Question

Đã trả lời bởi chuyên gia

Answer 1

$b ạ n x e m l ạ i s ố l i ệ u b à i x e m, s ố l i ệ u k h á l ớ n n ê n k ế t q u ả s ố c ũ n g t o G ọ i s ố m o l c ủ a C H_{4}, C_{2} H_{4}, C_{2} H_{2} l ầ n l ư ợ t l à x, y, z . m_{h ỗ n h ợ p} = 20, 8 g \Rightarrow 16 x + 28 y + 26 z = 20, 8 (1) * m_{H_{2} S O_{4} t r c' p / ư} = \frac{C % . m_{d d}}{100 %} = \frac{98.100}{100} = 98 g a m * k h ố i l ư ợ n g a x i t t r ư ớ c v à s a u p h ả n ứ n g k h ô n g đ ổ i n ê n m_{H_{2} S O_{4} t r c' p / ư} = m_{H_{2} S O_{4} s a u p / ư} = 98 g a m * m_{d d H_{2} S O_{4} s a u p / ư} = \frac{m_{c t}}{C %} . 100 % = \frac{98}{78, 27} . 100 \approx 125, 21 g a m \Rightarrow m_{b ì n h d d H_{2} S O_{4} t ă n g} = 125, 21 - 100 = 25, 21 g a m = m_{H_{2} O} * B T N T (H) c ó : 18 . (2 x + 2 y + z) = 25, 21 (2) * n_{h ỗ n h ợ p} = V : 22, 4 = 5, 04 : 22, 4 = 0, 225 m o l = x + y + z n_{B r_{2}} = \frac{C % . m_{d d}}{100 . M} = \frac{8.500}{100.160} = 0, 25 m o l = n_{C_{2} H_{4}} + n_{C_{2} H_{2}} = y + 2 z \Rightarrow \frac{x + y + z}{y + 2 z} = \frac{0, 225}{0, 25} \Rightarrow 0, 25 x + 0, 025 y - 0, 2 z = 0 (3) * G i ả i h ệ (1), (2), (3) t a ư ợ c : x \approx 0, 288 m o l; y \approx 0, 218 m o l; z \approx 0, 388 m o l a / % m_{C H_{4}} = \frac{0, 288.16}{20, 8} . 100 \approx 22, 15 % % m_{C_{2} H_{4}} = \frac{0, 218.28}{20, 8} . 100 \approx 29, 35 % % m_{C_{2} H_{2}} = 100 - 22, 15 - 29, 35 = 48, 5 % b / C H_{4} + 2 O_{2} \overset{t 0}{\to} C O_{2} + 2 H_{2} O (4) x m o l . . . 2 x m o l C_{2} H_{4} + 3 O_{2} \overset{t 0}{\to 2} C O_{2} + 2 H_{2} O (5) y m o l . . . . 3 y m o l C_{2} H_{2} + \frac{5}{2} O_{2} \overset{t 0}{\to 2} C O_{2} + H_{2} O (6) z m o l . . . . . . \frac{5 z}{2} m o l \sum_{} n_{O_{2}} = 2 x + 3 y + \frac{5 z}{2} = 2.0, 288 + 3.0, 218 + \frac{5.0, 388}{2} = 2, 2 m o l 20, 8 g a m h h c ầ n 2, 2 m o l O_{2} đ ể đ ố t c h á y v ậ y 200 g a m h h c ầ n \approx 21, 15 m o l O_{2} \Rightarrow n_{k k} = \frac{n_{O_{2}}}{20 %} = \frac{21, 15}{20} . 100 = 105, 75 m o l \Rightarrow V_{k k} = n_{k k} . 22, 4 = 105, 75.22, 4 = 2368, 8 l í t c / C O_{2} + C a {(O H)}_{2} \to C a C O_{3} + H_{2} O n_{C a {(O H)}_{2}} = n_{C a C O_{3}} = C_{M} . V = 0, 025.40 = 1 m o l m_{C a C O_{3}} = n . M = 1.100 = 100 g a m$

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Answer 2

$b ạ n x e m l ạ i s ố l i ệ u b à i x e m, s ố l i ệ u k h á l ớ n n ê n k ế t q u ả s ố c ũ n g t o G ọ i s ố m o l c ủ a C H_{4}, C_{2} H_{4}, C_{2} H_{2} l ầ n l ư ợ t l à x, y, z . m_{h ỗ n h ợ p} = 20, 8 g \Rightarrow 16 x + 28 y + 26 z = 20, 8 (1) * m_{H_{2} S O_{4} t r c' p / ư} = \frac{C % . m_{d d}}{100 %} = \frac{98.100}{100} = 98 g a m * k h ố i l ư ợ n g a x i t t r ư ớ c v à s a u p h ả n ứ n g k h ô n g đ ổ i n ê n m_{H_{2} S O_{4} t r c' p / ư} = m_{H_{2} S O_{4} s a u p / ư} = 98 g a m * m_{d d H_{2} S O_{4} s a u p / ư} = \frac{m_{c t}}{C %} . 100 % = \frac{98}{78, 27} . 100 \approx 125, 21 g a m \Rightarrow m_{b ì n h d d H_{2} S O_{4} t ă n g} = 125, 21 - 100 = 25, 21 g a m = m_{H_{2} O} * B T N T (H) c ó : 18 . (2 x + 2 y + z) = 25, 21 (2) * n_{h ỗ n h ợ p} = V : 22, 4 = 5, 04 : 22, 4 = 0, 225 m o l = x + y + z n_{B r_{2}} = \frac{C % . m_{d d}}{100 . M} = \frac{8.500}{100.160} = 0, 25 m o l = n_{C_{2} H_{4}} + n_{C_{2} H_{2}} = y + 2 z \Rightarrow \frac{x + y + z}{y + 2 z} = \frac{0, 225}{0, 25} \Rightarrow 0, 25 x + 0, 025 y - 0, 2 z = 0 (3) * G i ả i h ệ (1), (2), (3) t a ư ợ c : x \approx 0, 288 m o l; y \approx 0, 218 m o l; z \approx 0, 388 m o l a / % m_{C H_{4}} = \frac{0, 288.16}{20, 8} . 100 \approx 22, 15 % % m_{C_{2} H_{4}} = \frac{0, 218.28}{20, 8} . 100 \approx 29, 35 % % m_{C_{2} H_{2}} = 100 - 22, 15 - 29, 35 = 48, 5 % b / C H_{4} + 2 O_{2} \overset{t 0}{\to} C O_{2} + 2 H_{2} O (4) x m o l . . . 2 x m o l C_{2} H_{4} + 3 O_{2} \overset{t 0}{\to 2} C O_{2} + 2 H_{2} O (5) y m o l . . . . 3 y m o l C_{2} H_{2} + \frac{5}{2} O_{2} \overset{t 0}{\to 2} C O_{2} + H_{2} O (6) z m o l . . . . . . \frac{5 z}{2} m o l \sum_{} n_{O_{2}} = 2 x + 3 y + \frac{5 z}{2} = 2.0, 288 + 3.0, 218 + \frac{5.0, 388}{2} = 2, 2 m o l 20, 8 g a m h h c ầ n 2, 2 m o l O_{2} đ ể đ ố t c h á y v ậ y 200 g a m h h c ầ n \approx 21, 15 m o l O_{2} \Rightarrow n_{k k} = \frac{n_{O_{2}}}{20 %} = \frac{21, 15}{20} . 100 = 105, 75 m o l \Rightarrow V_{k k} = n_{k k} . 22, 4 = 105, 75.22, 4 = 2368, 8 l í t c / C O_{2} + C a {(O H)}_{2} \to C a C O_{3} + H_{2} O n_{C a {(O H)}_{2}} = n_{C a C O_{3}} = C_{M} . V = 0, 025.40 = 1 m o l m_{C a C O_{3}} = n . M = 1.100 = 100 g a m$

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