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Helpme29382
@Helpme29382
August 2022
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3г оксида лития растворили в 54 г воды.Найти массовую долю щелочи в образовавшемся растворе
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Alexei78
Дано
m(Li2O) = 3 g
m(H2O) = 54 g
----------------------
W(LiOH)-?
Li2O+H2O-->2LiOH
M(Li2O) = 30 g/mol
n(Li2O) = m/M = 3/30 = 0.1 mol
n(Li2O)= 2n(LiOH)
n(LiOH) = 2*0.1 = 0.2 mol
M(LiOH) = 24 g/mol
m(LiOH) = n*M= 0.2*24 =4.8 g
W(LiOH) = 4.8 / ( 3+54) * 100% = 8.42%
ответ 8.42%
2 votes
Thanks 5
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Answers & Comments
m(Li2O) = 3 g
m(H2O) = 54 g
----------------------
W(LiOH)-?
Li2O+H2O-->2LiOH
M(Li2O) = 30 g/mol
n(Li2O) = m/M = 3/30 = 0.1 mol
n(Li2O)= 2n(LiOH)
n(LiOH) = 2*0.1 = 0.2 mol
M(LiOH) = 24 g/mol
m(LiOH) = n*M= 0.2*24 =4.8 g
W(LiOH) = 4.8 / ( 3+54) * 100% = 8.42%
ответ 8.42%