Ba(NO3)2 + Na2SO4 = BaSO4 + 2NaNO3
m р-ра (Ba(NO3)2) = 120 г
W = 10%
m(BaSO4) = ?
m(Ba(NO3)2) = 120 * 0.1 = 12 г
n(Ba(NO3)2) = 12/261 = 0.04 моль
n(Ba(NO3)2) = n(BaSO4) = 0.04 моль
m(BaSO4) = Mn = 0.04 * 233 = 9.32 г
Ответ: 9.32 г
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Ba(NO3)2 + Na2SO4 = BaSO4 + 2NaNO3
m р-ра (Ba(NO3)2) = 120 г
W = 10%
m(BaSO4) = ?
m(Ba(NO3)2) = 120 * 0.1 = 12 г
n(Ba(NO3)2) = 12/261 = 0.04 моль
n(Ba(NO3)2) = n(BaSO4) = 0.04 моль
m(BaSO4) = Mn = 0.04 * 233 = 9.32 г
Ответ: 9.32 г