find the amount of MG in grams to be dissolved in dilute H2 S o4 to liberate H2 which is just sufficient to reduce 160 grams of ferric oxide is
Answers
The equation for the reduction of ferric oxide (Fe₂O₃) by H₂ :
Fe₂O₃ + 3H₂ → 2Fe + 3H₂O
Mole ratio Fe₂O₃ : H₂ = 1 : 3
Molar mass of Fe₂O₃ = (55.8×2 + 16.0×3) g/mol = 159.6 g/mol
No. of moles of Fe₂O₃ reacted = (160 g) / (159.6 g/mol) = 1.00 mol
No. of moles of H₂ reacted = (1.00 mol) × 3 = 3.00 mol
The equation for the reaction between Mg and dilute H₂SO₄ :
Mg + H₂SO₄ → MgSO₄ + H₂
Mole ratio Mg : H₂ = 1 : 1
No. of moles of H₂ formed = 3.00 mol
No. of moles of Mg needed = (3.00 mol) × 1 = 3.00 mol
Molar mass of Mg = 24.3 g/mol
Mass of Mg needed = (3.00 mol) × (24.3 g/mol) = 72.9 g
Answer:
Explanation:the reaction between fe203 and h2
Fe2o3+3h2 gives rise to 2fe +3h20
So here for 160 g fe2o3 6g of h2 is used
Now the same 6g of h2 is used in the below equation as per the question
Mg+h2so4 gives rise to mgso4 +h2
So
24 g of mg requires 2g of h2 as per equation and we need to find for 6g of h2 which is.
6*24/2=72 g of mg