1. Maria reacts 1.0422 g of C7H6O3 (138.137 g/mol) and 0.9045 g of C4H8O3 (102.104 g/mol) to produce 0.8664 g of C₂H8O4

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answerhappygod
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1. Maria reacts 1.0422 g of C7H6O3 (138.137 g/mol) and 0.9045 g of C4H8O3 (102.104 g/mol) to produce 0.8664 g of C₂H8O4

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1 Maria Reacts 1 0422 G Of C7h6o3 138 137 G Mol And 0 9045 G Of C4h8o3 102 104 G Mol To Produce 0 8664 G Of C H8o4 1
1 Maria Reacts 1 0422 G Of C7h6o3 138 137 G Mol And 0 9045 G Of C4h8o3 102 104 G Mol To Produce 0 8664 G Of C H8o4 1 (39.5 KiB) Viewed 46 times
1. Maria reacts 1.0422 g of C7H6O3 (138.137 g/mol) and 0.9045 g of C4H8O3 (102.104 g/mol) to produce 0.8664 g of C₂H8O4 (180.179 g/mol) and acetic acid. What are the theoretical and percent yields for the reaction? (Hint: first determine limiting reagent.) C7H6O3 + C4H6O3C9H8O4 + C₂H4O2 2. Robin is tasked with synthesizing p-nitrotoluene from toluene in excess concentrated nitric acid. She starts with 50.00 mL of toluene (92.141 g/mol, d=0.866 g/mL), and obtains 56.0662 g of p-nitrotoluene (137.138 g/mol). What is her percent yield? C7H8+ HNO3 → C7H7NO2 + H₂O
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