Question

When 0.5 mols of colemanite $\left(C{a}_2{B}_6{O}_{11}\right)$ is treated with $S{O}_2$ in aqueous solution to form orthoboric acid, how many grams of orthoboric acid will be formed?
(Take molar mass of boron 11 gm/mol, oxygen 16 gm/mol)

Answer 1

$C{a}_2{B}_6{O}_{11}+4S{O}_2+11H_{2}O\to 2Ca(HS{O}_3{)}_2+6H_{3}B{O}_3$
0.5 mol of colemanite will produce $0.5\times 6=3$ mols of $H_{3}B{O}_3$
Mass of $H_{3}B{O}_3=3\times 62=186gmof{H}_{3}B{O}_3$