In a laboratory preparation of triethyl borate for the green-flame test, 0.25 mol of borax (Na2B4O7) is treated with dilute sulphuric acid to liberate boric acid, Na2B4O7 + H2SO4 + 5H2O -> 4H3BO3 + Na2SO4, which is then esterified, H3BO3 + 3C2H5OH -> B(OC2H5)3 + 3H2O. The minimum amount of ethanol needed for complete esterification is

JEE Advanced levelSingle correct optionDifficulty 3 of 3qa074

In a laboratory preparation of triethyl borate for the green-flame test, $0.25 mol$ of borax ( $Na X_{2} B X_{4} O X_{7}$ ) is treated with dilute sulphuric acid to liberate boric acid, $Na X_{2} B X_{4} O X_{7} + H X_{2} SO X_{4} + 5 H X_{2} O 4 H X_{3} BO X_{3} + Na X_{2} SO X_{4}$ , which is then esterified, $H X_{3} BO X_{3} + 3 C X_{2} H X_{5} OH B (OC X_{2} H X_{5}) X_{3} + 3 H X_{2} O$ . The minimum amount of ethanol needed for complete esterification is

0.25 mol

of borax gives

4 \times 0.25 = 1.0 mol

H X_{3} BO X_{3}

. Each

H X_{3} BO X_{3}

needs 3 ethanol, so ethanol

= 3 \times 1.0 = 3.0 mol