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Helpme29382
@Helpme29382
July 2022
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6,2 г оксида натрия растворили в 100 г воды. Найти массовую долю щелочи в образовавшемся растворе.
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Alexei78
Дано
m(Na2O) = 6.2 g
m(H2O) = 100 g
-------------------------
W(NaOH) -?
Na2O+H2O-->2NaOH
M(Na2O) = 62 g/mol
n(Na2O) = m/M = 6.2/62 = 0.1 mol
n(Na2O) = 2n(NAOH) = 0.1*2 = 0.2 mol
M(NaOH) = 40 g/mol
m(NaOH) = n*M = 0.2*40 = 8 g
W(NaOH) = 8 / (6.2+100) *100% = 7.54 %
ответ 7.54 %
2 votes
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Answers & Comments
m(Na2O) = 6.2 g
m(H2O) = 100 g
-------------------------
W(NaOH) -?
Na2O+H2O-->2NaOH
M(Na2O) = 62 g/mol
n(Na2O) = m/M = 6.2/62 = 0.1 mol
n(Na2O) = 2n(NAOH) = 0.1*2 = 0.2 mol
M(NaOH) = 40 g/mol
m(NaOH) = n*M = 0.2*40 = 8 g
W(NaOH) = 8 / (6.2+100) *100% = 7.54 %
ответ 7.54 %