Question

Neodymium battery + sulfuric acid ——> neodymium sulfate + iron sulfate + boron + hydrogen. I have 5g of Nd2Fe14B and enough H2SO4, how much NdSO4 will I get?

Answer 1

Mass of Nd₂(SO₄)₃ : 2.653 g

Further explanation

Reaction

Nd₂Fe₁₄B + 17H₂SO₄ → Nd₂(SO₄)₃ + 14FeSO₄ + B + 17H₂

5 g Nd₂Fe₁₄B

MW = 2. 144,242 + 14. 55.845 + 10.811

MW = 288.484 + 781.83+10.811=1081.125 g/mol

mol :

`\tt mol=(mass)/(MW)\\\\mol=(5)/(1081.125)=0.0046`

mol Nd₂Fe₁₄B = mol Nd₂(SO₄)₃=0.0046

so mass of Nd₂(SO₄)₃ :

MW Nd₂(SO₄)₃ = 2.144.242+3.32,065+12.15,999

MW = 288.484+96.195+191.988=576.667 g/mol

`\tt mass=mol* MW\\\\mass=0.0046* 576.667\\\\mass=2.653~g`