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What is the maximum number of grams of B2H6(g) that can be prepared from 2.650g of NaBH4(s) and 4.560g of BF3(g)?Equation for Reaction: 3 NaBH4(s) 4 BF3(g) --> 3 NaBF4(s) 2 B2H6(g)--Thank you so much.
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