Br 8 NH₂ 9 HC(OME)3 reflux T=" Ph₂O reflux H Br 10 Br 11 2,2-Dimethyl-5-(4-bromophenylaminomethylene)-1,3-dioxan-4,6-dione (10): Dissolve Meldrum's acid (9, pKa = 4.97) (4.60 g, 31.9 mmol) in methyl orthoformate (60 mL) and then heat to reflux for 2 h under a nitrogen atmosphere (use a balloon). Cool the reaction mixture to room temperature then add 4-bromoaniline (3.80 g, 22.1 mmol) and heat to reflux for 6 h.5 Allow to cool to room temperature and collect the precipitated solid and recrystallize from methanol. Obtain a melting point for the resulting purified solid. 6-Bromo-4-oxo-4H-quinoline (11): Dissolve 10 (2.00 g, 6.12 mmol) in hot6 diphenyl ether (31 mL) and then heat to reflux until the formation of gaseous products ceases. After cooling to room temperature, add an equal volume of petroleum ether (bp 40- 60 °C) and allow to stand for 48 h. Isolate the product by filtration and wash with petroleum ether.7 Use deuterated DMSO as the solvent when preparing an NMR sample of this compound.

[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.

The actual yield of one of the reactions in 43.5g. What was theoretical yield in the laboratory if the scientist concluded that % yield of the reaction ended up to be 77%.

For the reaction of sodium sulfide with iron(III) chloride below (as an unbalanced reaction - that's how I will almost always give you reactions.  ALWAYS check to see if a reaction is balanced before you use it.  You balanced it in the last question; make sure you have that right before you start calculating!), you have 319.5 kg of sodium sulfide and 502.3 kg of iron(III) chloride available. What mass (in kilograms) of sodium sulfide do you NEED if you're going to reduce  all of the iron(III) chloride you have?   Na2S(s)  +   FeCl3(s)    →   Fe2S3(s)   +  NaCl(s)