Question

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Answer 1

$a, n_{{Na}_{2} O} = 3, 1 : 62 = 0, 05 mol {Na}_{2} O + H_{2} O - > 2 NaOH 0, 05 - - - - - > 0, 1 (mol) b, C_{M NaOH} = 0, 1 : 0, 15 = \frac{2}{3} M c, 2 NaOH + H_{2} {SO}_{4} - > {Na}_{2} {SO}_{4} + 2 H_{2} O 0, 1 - - - > 0, 05 (mol) m_{H_{2} {SO}_{4}} = 0, 05.98 = 4, 9 g m_{dd H_{2} {SO}_{4}} = 4, 9 : 0, 2 = 24, 5 g V_{dd H_{2} {SO}_{4}} = 24, 5 : 1, 14 = 21, 49 ml$

vậy V=21,49 ml

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Answer 2

$a, n_{{Na}_{2} O} = 3, 1 : 62 = 0, 05 mol {Na}_{2} O + H_{2} O - > 2 NaOH 0, 05 - - - - - > 0, 1 (mol) b, C_{M NaOH} = 0, 1 : 0, 15 = \frac{2}{3} M c, 2 NaOH + H_{2} {SO}_{4} - > {Na}_{2} {SO}_{4} + 2 H_{2} O 0, 1 - - - > 0, 05 (mol) m_{H_{2} {SO}_{4}} = 0, 05.98 = 4, 9 g m_{dd H_{2} {SO}_{4}} = 4, 9 : 0, 2 = 24, 5 g V_{dd H_{2} {SO}_{4}} = 24, 5 : 1, 14 = 21, 49 ml$

vậy V=21,49 ml

Answer 3

a)

$PTHH : {Na}_{2} O + H_{2} O \to 2 NaOH$

b)

$n_{{Na}_{2} O} = \frac{3, 1}{23.2 + 16} = 0, 05 (mol) \Rightarrow n_{NaOH} = 2 . n_{{Na}_{2} O} = 2.0, 05 = 0, 1 (mol) \Rightarrow C_{M_{dd NaOH}} = \frac{n}{V} = \frac{0, 1}{0, 15} = \frac{2}{3} (M)$

c)

$2 NaOH + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + 2 H_{2} O \Rightarrow n_{H_{2} {SO}_{4}} = \frac{1}{2} . n_{NaOH} = \frac{1}{2} . 0, 1 = 0, 05 (mol) \Rightarrow m_{H_{2} {SO}_{4}} = 0, 05.98 = 4, 9 (g) \Rightarrow m_{dd H_{2} {SO}_{4}} = \frac{m_{H_{2} {SO}_{4}}}{C %} = \frac{4, 9}{20 %} = 24, 5 (g) \Rightarrow V_{dd H_{2} {SO}_{4}} = \frac{m_{dd H_{2} {SO}_{4}}}{D} = \frac{24, 5}{1, 14} = 21, 49 (ml)$

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Answer 4

a)

$PTHH : {Na}_{2} O + H_{2} O \to 2 NaOH$

b)

$n_{{Na}_{2} O} = \frac{3, 1}{23.2 + 16} = 0, 05 (mol) \Rightarrow n_{NaOH} = 2 . n_{{Na}_{2} O} = 2.0, 05 = 0, 1 (mol) \Rightarrow C_{M_{dd NaOH}} = \frac{n}{V} = \frac{0, 1}{0, 15} = \frac{2}{3} (M)$

c)

$2 NaOH + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + 2 H_{2} O \Rightarrow n_{H_{2} {SO}_{4}} = \frac{1}{2} . n_{NaOH} = \frac{1}{2} . 0, 1 = 0, 05 (mol) \Rightarrow m_{H_{2} {SO}_{4}} = 0, 05.98 = 4, 9 (g) \Rightarrow m_{dd H_{2} {SO}_{4}} = \frac{m_{H_{2} {SO}_{4}}}{C %} = \frac{4, 9}{20 %} = 24, 5 (g) \Rightarrow V_{dd H_{2} {SO}_{4}} = \frac{m_{dd H_{2} {SO}_{4}}}{D} = \frac{24, 5}{1, 14} = 21, 49 (ml)$

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