дано
m(ppa BaCL2) = 15 g
W(BaCL2) = 5%
m(ppa Na2SO4) = 10 g
W(Na2SO4)= 8%
m(BaSO4)-?
m(BaCL2) = 15*5% / 100% = 0.75 g
M(BaCL2) = 208 g/mol
n(BaCL2) = m/M = 0.75 / 208 = 0.0036 mol
m(Na2SO4) = 10*8% / 100% = 0.8g
M(Na2SO4) = 142 g/mol
n(Na2SO4) = m/M = 0.8 / 142 = 0.0056 mol
n(BaCL2) < n(Na2SO4)
BaCL2+Na2SO4-->2NaCL+BaSO4
n(BaCL2) = n(BaSO4) = 0.0036 mol
M(BaSO4) = 233g/mol
m(BaSO4) = n*M = 0.0036 * 233 = 0.8388 g
ответ 0.8388 г
дано
m(ppa BaCL2) = 15 g
W(BaCL2) = 5%
m(ppa Na2SO4) = 10 g
W(Na2SO4)= 8%
m(BaSO4)-?
m(BaCL2) = 15*5% / 100% = 0.75 g
M(BaCL2) = 208 g/mol
n(BaCL2) = m/M = 0.75 / 208 = 0.0036 mol
m(Na2SO4) = 10*8% / 100% = 0.8g
M(Na2SO4) = 142 g/mol
n(Na2SO4) = m/M = 0.8 / 142 = 0.0056 mol
n(BaCL2) < n(Na2SO4)
BaCL2+Na2SO4-->2NaCL+BaSO4
n(BaCL2) = n(BaSO4) = 0.0036 mol
M(BaSO4) = 233g/mol
m(BaSO4) = n*M = 0.0036 * 233 = 0.8388 g
ответ 0.8388 г