Answer:
52.5 mol O2
Explanation:
4 FeCl3 + 6 O2 -> 2 Fe2O3 +6 Cl2
4 mol FeCl3 -> 6 mol O2
35.0 mol FeCl3 -> x
x= (35.0 mol FeCl3 * 6 mol O2)/4 mol FeCl3
x=52.5 mol O2
Answer:
Explanation:
Number of moles of CuCl2 initially present = volume * molar concentration
= 0.134 * 8
= 1.072 mol.
Molar mass of CuCl2 = 63.5 + (2*35.5)
= 134.5 g/mol
Mass of CuCl2 = molar mass * number of moles
= 134.5 * 1.072
= 144.184 g
Mass of CuCl2 in 48 ml = 5.89 g in 48 ml
Volume = 5.89 * (48/144.184)
= 1.96 ml.
Its FeSO3 or iron(iii)sulfite = Fe2(SO3)3
