How many moles of aluminum are needed to react completely with 4.8 mol of FeO? 2Al(s) + 3FeO(s)- 3c(s) + ABO3()

How many moles of aluminum are needed to react completely with 1.2 mol of FeO? 2Al(s) + 3FeO(s) = 3Fe(s) + Al2O3(s). 0.8 mols.

Explanation:

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm