Reduction of boron trichloride by hydrogen reaction
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Answer:
2BCl3 + 3H2 → 6HCl + 2B
or BCl3 + 3/2H2 → 3HCl + B
10.8 g boron requires hydrogen = 3/2 22.4 L
21.6 g boron will require hydrogen
= 3x22.4/(2x10.8) = 67.2 L
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