дано
m(ppa BaCL2) = 200 g
W(BaCL2) = 10%
m(AgCL)-?
m(BaCL2) = 200 * 10% / 100% = 20 g
BaCL2+2AgNO3-->Ba(NO3)2+2AgCL
M(BaCL2) = 208 g/mol
n(BaCL2) = m/M = 20 / 208 = 0.096 mol
n(BaCL2) = 2n(AgCL)
n(AgCL) = 0.096*2 = 0.192 mol
M(AgCL) = 172.5 mol
m(AgCL) = (AgCL) * M(AgCL) = 0.192 * 172.5 = 33.12 g
ответ 33.12 г
Объяснение:
дано
m(ppa BaCL2) = 200 g
W(BaCL2) = 10%
m(AgCL)-?
m(BaCL2) = 200 * 10% / 100% = 20 g
BaCL2+2AgNO3-->Ba(NO3)2+2AgCL
M(BaCL2) = 208 g/mol
n(BaCL2) = m/M = 20 / 208 = 0.096 mol
n(BaCL2) = 2n(AgCL)
n(AgCL) = 0.096*2 = 0.192 mol
M(AgCL) = 172.5 mol
m(AgCL) = (AgCL) * M(AgCL) = 0.192 * 172.5 = 33.12 g
ответ 33.12 г
Объяснение: