Thermochemistry

Question

3 MgO + 2 Fe --> 2 Mg + Fe₂O₃        ΔHrxn=-47,524 kJ/mol_rxnThe reaction above is utilized by campers to recycle their drinking water by distilling it (heating the water to 105^oC steam) and collecting the steam in a separate container to cool and ready to drink. The waste water tank holds the water at 25^oC. The tank itself has a mass of 12g with a specific heat of 3.17J/g^oC - the melting point of the plastic is well above 105^oC. Assume that the water and tank are at the same initial temperature. If the camper contains 192,276g of iron, how much water will they be able to distill?

Answer 1

3 MgO + 2Fe --> 3 Mg + Fe₂O₃ ... ∆H_rxn = -47,524 kJ/mol

moles of Fe available = 192,276 g Fe x 1 mol Fe / 55.85 g = 3443 mols Fe

Heat available = 3443 mols Fe x 47,524 kJ / mol = 1.64x10⁸ kJ

q = m_H2O x C_H2O x ∆T_H2O + m_tank x C_tank x ∆T_tank + ∆H_vap x m_H2O + m_H2O x C_steam x ∆T

1.64x10⁸ kJ = (m_H2O)(4.184 kJ/kgº)(75º) + (0.012 kg)(3.17 kJ/kgº)(75º) + 2260 kJ/kg(m_H2O) + (m_H2O)(1.89 kJ/kgº)(5º)

1.64x10⁸ kJ = 314m_H2O + 2.85 + 2260m_H2O + 9.45m_H2O

1.64x10⁸ kJ = 252m_H2O

m_H2O = 6.51x10⁵ kg of H2O

(please be sure to check the math)