Verified answer

Fe2O3+3CO=2Fe+3CO2

w(Fe2O3)=100%-10%=90%=0,9

m(Fe2O3)=960*0,9=864г

M(Fe2O3)=160г/моль

M(Fe)=56г/моль

n(Fe2O3)=864/160=5,4моль

nFe2O3:nFe=1:2

n(Fe)=5,4*2=10,8моль

m(Fe)=10,8*56=604,8г

Verified answer

дано

m(Fe2O3) = 960 g

W(прим) = 10%

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m(Fe) -?

m чист (Fe2O3) = 960 - (960 * 10% / 100% ) = 864 g

Fe2O3+3CO-->2Fe+3CO2

M(Fe2O3) = 160 g/mol

n(Fe2O3) = m/M = 864 / 160 == 5.4 mol

n(Fe) = 2n(Fe)

n(Fe) = 5.4*2 = 10.8 mol

M(Fe) = 56 g/mol

m(Fe) = n(Fe)*M(Fe) = 10.8*56 = 604.8 g

ответ 604.8 г