Given 7.75 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield? express your answer in grams to three significant fi

10.2 grams of ethyl butyrate synthesized. The balanced equation for the reaction of butanoic acid (C4H8O2) with ethanol (C2H6O) to produce ethyl butyrate (C6H12O2) is: C4H8O2 + C2H6O ==> C6H12O2 + H2O So for each mole of C4H8O2 used, 1 mole of C6H12O2 will be produced. So let's calculate the reactant and product molar masses. Start by looking up the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass C4H8O2 = 4 * 12.0107 + 8 * 1.00794 + 2 * 15.999 = 88.10432 g/mol Molar mass C6H12O2 = 6 * 12.0107 + 12 * 1.00794 + 2 * 15.999 = 116.15748 g/mol Moles C4H8O2 = 7.75 g / 88.10432 g/mol = 0.087963905 mol Mass C6H12O2 = 0.087963905 mol * 116.15748 g/mol = 10.21766549 g Rounding to 3 significant figures gives 10.2 grams of ethyl butyrate synthesized.

Sorry i dont understand the question can you rephase it?