1) FeCO3+2NaOH=Fe(OH)2+Na2CO3
2) 4Fe(OH)2+O2=2Fe2O3+4H2O
M(Fe2O3)=159.6900 g/mol
n (Fe2O3)=m/M=0.2864/159.6900=0.0018 mol
In the second reaction: n (Fe(OH)2)=2 x n (Fe2O3)=0.0018 x 2 = 0.0036 mol
In the first reaction: n (FeCO3) = n (Fe(OH)2) = 0.0036 mol
M (FeCO3) = 115.8540 g/mol
m (FeCO3) = M x n = 115.8540 x 0.0036 = 0.4156 g
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