1) Boron will be oxidized with concentrated sulfuric acid up to boric acid formation:
B + H2SO4 = H3BO3 + SO2 + H2O
SO42- + 4H+ + 2e = SO2 + 2H2O
B + 3H2O - 3e = BO33- + 6H+
Thus,
2B + 3H2SO4 = 3SO2 + 2H3BO3
2) Lithium tetrahydroborate will be completely hydrolised and the products will react to form lithium borate and hydrogen gas:
LiBH4 + 2H2O = LiBO2 + 4H2
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