The following equation represents the reaction between aluminum (Al) and bromine (Br2) to produce aluminum bromide (AlBr3): 2Al  +  3 Br2  →  2AlBr3 If I try to react 15 grams of aluminum (Al) with 15 grams of bromine (Br2), calculate the number of moles of each reactant, and based on your answers, state which is the limiting reactant, Al or Br2. For your moles, give your answer to 3 significant figures.

[Tutorial: Limiting reactant stoichiometry] This question will walk you through the steps of calculating the mass of products produced based on your determination of the limiting reactant.   b) Step 2a: Use dimensional analysis to determine the theoretical yield of the product. Calculate the theoretical yield in grams Al₂O₃ from the complete reaction of 64.7 grams Al according to the following balanced chemical equation: 2 Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2 Fe(s) c) Calculate the theoretical yield in grams Al₂O₃ from the complete reaction of 201 grams Fe₂O₃ according to the following balanced chemical equation: 2 Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2 Fe(s) d) Which of the following substances is the limiting reactant? e) What is the mass in grams of the excess Fe₂O₃ remaining after the partial reaction of 201 g Fe₂O₃ with 64.7 g Al? Give your answer to three significant figures.

How much (in gram) of HCl is required to produce 21.45 g of  AlCl3 if the percentage yield for the following reaction is 61.24% ? Enter your answer without units. Molar masses (in g/mol):Al = 26.98H2 = 2.016HCl = 36.46AlCl3 = 133.34 2Al(s) + 6 HCl (aq)  ⟶⟶  2 AlCl3(s) + 3H2(g)

Consider the following reaction between sulfur trioxide and water: SO3(g)+H2O(l)→H2SO4(aq) A chemist allows 61.5 g of SO3SO3 and 11.2 g of H2O to react. When the reaction is finished, the chemist collects 56.2 g of H2SO4 Determine the percent yield for the reaction.         Express your answer to three significant figures.