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BCl3 +LiAlH4 -> B2H6 + LiAlCl4 calculate the moles of BCl3 that will be required to produce 985grams of B2H6

BCl3 +LiAlH4 -> B2H6 + LiAlCl4

calculate the moles of BCl3 that will be required to produce 985grams of B2H6

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sol Given mass of B H 6 Molar mass of BH 27 moles BH6 35611 LI XI C 11 985 ... blur-text-image

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