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sneginka22
@sneginka22
August 2022
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Сколько г СаСО3 выпадает в осадок при смешении 30 г 10 %-ного раствора СаСl2 и 40 мл 0,6 М раствора Na2СО3
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Гарда11
Na2CO3 + CaCl2 = CaCO3 + 2NaCl
m ( CaCl2) = 30*0.1 = 3 г
n (CaCl2) = m/M = 3/111 = 0.027 моль
n (Na2CO3) = 0.6*40/1000 = 0.024 моль
n (CaCl2) > n (Na2CO3) ⇒ n (CaCO3) = 0.024 моль
m (CaCO3) = 0.024*100 = 24 г
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Answers & Comments
m ( CaCl2) = 30*0.1 = 3 г
n (CaCl2) = m/M = 3/111 = 0.027 моль
n (Na2CO3) = 0.6*40/1000 = 0.024 моль
n (CaCl2) > n (Na2CO3) ⇒ n (CaCO3) = 0.024 моль
m (CaCO3) = 0.024*100 = 24 г