дано
m(ppa NaOH) = 100 g
w(NaOH) = 5%
m(ppa FeCL2) = 200 g
w(FeCL2) = 2%
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m(Fe(OH)2)-?
m(NaOH) = 100*5% / 100% = 5 g
M(NaOH) = 40 g/mol
n(NaOH) = m/M = 5 / 40 = 0.125 mol
m(FeCL2) = 200*2% / 100% = 4 g
M(FeCL2) = 127 g/mol
n(FeCL2) = m/M = 4/ 127 = 0.031 mol
n(NaOH) > n(FeCL2)
2NaOH+FeCL2 --> Fe(OH)2+2NaCL
M(Fe(OH)2) = 90 g/mol
n(FeCL2) = n(Fe(OH)2) = 0.031 mol
m(Fe(OH)2) = n*M = 0.0*90 = 2.79 g
ответ 2.79 г