дано
mppa(H2SO4) = 120 g
W(H2SO4) = 16%
m(BaSO4)-?
m(H2SO4) = 120 * 16% / 100% = 19.2 g
H2SO4+BaCL2-->2HCL+BaSO4
M(H2SO4) = 98 g/mol
n(H2SO4) = m/M = 19.2 / 98 = 0.2 mol
n(H2SO4) = n(BaSO4) = 0.2 mol
M(BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.2 * 233 = 46.6 g
ответ 46.6 г
Объяснение:
дано
mppa(H2SO4) = 120 g
W(H2SO4) = 16%
m(BaSO4)-?
m(H2SO4) = 120 * 16% / 100% = 19.2 g
H2SO4+BaCL2-->2HCL+BaSO4
M(H2SO4) = 98 g/mol
n(H2SO4) = m/M = 19.2 / 98 = 0.2 mol
n(H2SO4) = n(BaSO4) = 0.2 mol
M(BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.2 * 233 = 46.6 g
ответ 46.6 г
Объяснение: