if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced
? How many grams of each reagent would remain at the end of this reaction? what mass (in grams) of product would be collected if the reaction above3 proceeded in 72% yield?
2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g). n(Al) = 0,11 mol - 0,025 mol = 0,085 mol. m(Al) = 0,085 mol · 27 g/mol. m(Al) = 2,295 g. m(AlBr₃) = 6,67 g · 0,72 (yield of reaction). m(AlBr₃) = 4,8 g. n - amount of substance. M - molar mass.
An analysis of a 4.2-gram sample of a compound is found to contain 54.76 percent sodium (Na) and 45.24 percent fluorine (F). Determine the empirical formula of the compound?. <span>"NaF" </span> Hoped This Helped, <span>
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