Дано
m(ppa HCL) = 10 g
W(HCL)=60%
m(Fe)=8 g
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m(FeCL2)-?
m(HCL) = m(ppaHCL)*W(HCL)*100% = 10*60%/100% = 6 g
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 6/36.5 = 0.16 mol
M(Fe) = 56 g/mol
n(Fe) =m/M = 8/56 = 0.14 mol
n(HCL)>n(Fe)
8 X
Fe+2HCL-->FeCL2+H2 M(FeCL2)= 127 g/mol
56 127
X= 8*127 / 56 = 18.14 g
ответ 18.14 г