Question

tinh khoi luong sat trong 50kg quang chua 80% fe2o3

Answer 1

m(Fe2O3)trong 50kgquặng

=50.(80/100)=40kg
=>% (Fe)trong Fe2O3=56.2/(56.2+16.3)

=0,7(tức 70%)
=>m(Fe)=40.0,7=28kg

Answer 2

m_Fe2O3 = 50.80% = 40 (kg)

%_{Fe trong Fe2O3} = \(\dfrac{56.2}{160}\)= 70%

⇒ m_Fe = 40.70% = 28 (kg)

Answer 3

*m(Fe2O3)trong 50kg quặng=50*(80/100)=40kg
=>%(Fe) trong Fe2O3=56*2/(56*2+16*3)=0,7(tức 70%)
=>m(Fe)=40*0,7=28kg