Дано:
n (BaSO4) = 0,1 моль
знайти:m (H2SO4), m (Ba (OH2))
рішення:
M (BaSO4) = 137 + 32 + 16 ∙ 4 = 233 г \ моль
m (BaSO4) = n ∙ M = 0,1 моль ∙ 233 г \ моль = 23,3 г
х г у г 23,3 г
H2SO4 + Ba(OH)2 = BaSO4 + H2O
98 г 171 г 233 г
Обчислимо масу сірчаної кислоти необхідної для реакції:
х г --- 23,3 г
98 г --- 233 г
х = 98 г ∙ 23,3 г \ 233 г = 9.8 г
Обчислимо масу гідроксиду барію необхідної для реакції:
у г --- 23,3 г
171 г --- 233 г
х = 171 г ∙ 23,3 г \ 233 г = 17,1 г
Відповідь: m (H2SO4) = 9,8 г, m (Ba (OH) 2) = 17,1 г
Объяснение:
Дур*
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Must Waste More Time...
ответ: P=P2O5=H2PO4=No3PO4=Aq3PO4=H2PO4
Объяснение: