Question

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Answer 1

$n_{A L} = 0, 06 m o l; n_{H_{2} S O_{4}} = 0, 3.0, 5 = 0, 15 m o l = > S a u p ư t h ì H_{2} S O_{4} d ư v à A l h ế t a, 2 A l + 3 H_{2} S O_{4} - > A l_{2} {(S O_{4})}_{3} + 3 H_{2} 0, 06 0, 09 0, 03 0, 09 (m o l) = > m_{A l_{2} {(S O_{4})}_{3}} = 0, 03.342 = 10, 26 g C_{M A l_{2} {(S O_{4})}_{3}} = \frac{0, 03}{0, 3} = 0, 1 M n_{H_{2} S O_{4}} d ư = 0, 15 - 0, 09 = 0, 06 m o l = > C_{M H_{2} S O_{4}} = \frac{0, 06}{0, 3} = 0, 2 M V ậ y : m_{A l_{2} {(S O_{4})}_{3}} = 10, 26 g; C_{M A l_{2} {(S O_{4})}_{3}} = 0, 1 M; C_{M H_{2} S O_{4}} d ư = 0, 2 M b, V_{H_{2}} = 0, 09.22, 4 = 2, 016 l í t V ậ y V = 2, 016 l í t$ .

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Answer 2

$n_{A L} = 0, 06 m o l; n_{H_{2} S O_{4}} = 0, 3.0, 5 = 0, 15 m o l = > S a u p ư t h ì H_{2} S O_{4} d ư v à A l h ế t a, 2 A l + 3 H_{2} S O_{4} - > A l_{2} {(S O_{4})}_{3} + 3 H_{2} 0, 06 0, 09 0, 03 0, 09 (m o l) = > m_{A l_{2} {(S O_{4})}_{3}} = 0, 03.342 = 10, 26 g C_{M A l_{2} {(S O_{4})}_{3}} = \frac{0, 03}{0, 3} = 0, 1 M n_{H_{2} S O_{4}} d ư = 0, 15 - 0, 09 = 0, 06 m o l = > C_{M H_{2} S O_{4}} = \frac{0, 06}{0, 3} = 0, 2 M V ậ y : m_{A l_{2} {(S O_{4})}_{3}} = 10, 26 g; C_{M A l_{2} {(S O_{4})}_{3}} = 0, 1 M; C_{M H_{2} S O_{4}} d ư = 0, 2 M b, V_{H_{2}} = 0, 09.22, 4 = 2, 016 l í t V ậ y V = 2, 016 l í t$ .

Answer 3

$n_{Al} = 0, 06 mol n_{H_{2} {SO}_{4}} = 0, 15 mol Có \frac{3}{2} n_{Al} < n_{H_{2} {SO}_{4}} = > H_{2} {SO}_{4} dư 2 Al + 3 H_{2} {SO}_{4} - > {Al}_{2} {({SO}_{4})}_{3} + 3 H_{2} 0, 06 0, 09 0, 03 0, 09 a, m_{{Al}_{2} {({SO}_{4})}_{3}} = 0, 03.324 = 9, 72 g C_{M {Al}_{2} {({SO}_{4})}_{3}} = 0, 03 : 0, 3 = 0, 1 M C_{M H_{2} {SO}_{4} dư} = (0, 15 - 0, 09) : 0, 3 = 0, 2 M b, V_{H_{2}} = 0, 09.22, 4 = 2, 016 lít$

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Answer 4

$n_{Al} = 0, 06 mol n_{H_{2} {SO}_{4}} = 0, 15 mol Có \frac{3}{2} n_{Al} < n_{H_{2} {SO}_{4}} = > H_{2} {SO}_{4} dư 2 Al + 3 H_{2} {SO}_{4} - > {Al}_{2} {({SO}_{4})}_{3} + 3 H_{2} 0, 06 0, 09 0, 03 0, 09 a, m_{{Al}_{2} {({SO}_{4})}_{3}} = 0, 03.324 = 9, 72 g C_{M {Al}_{2} {({SO}_{4})}_{3}} = 0, 03 : 0, 3 = 0, 1 M C_{M H_{2} {SO}_{4} dư} = (0, 15 - 0, 09) : 0, 3 = 0, 2 M b, V_{H_{2}} = 0, 09.22, 4 = 2, 016 lít$

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