Question

99PCT!Aflati masa de fier din 1712 g amestec FeO si Fe2O3, in care cele doua substante se gasesc in raport molar FeO:Fe2O3=3:4

Answer 1

m amestec = m FeO + mFe2O3
x - m FeO si y - m Fe2O3
x/y = 3/4 => 3y=4x => y = 4x/3
x+y = 1712
x + 4x/3 = 1712 
=> 3x + 4x = 5136 
=> x =  733.71 g FeO
y = 4x/3 = 4*733.71/3 = 978.28 g Fe2O3
70 g FeO........56 g Fe
733.71 g FeO...........a g Fe
a = 587 g Fe
160 g Fe2O3..........2*56 g Fe
978.28 g Fe2O3............b g Fe
b = 684.8 g Fe
masa Fe = a + b = 587+684.8 = 1271,8 g Fe