The equation $2\mathrm{Bi}(\mathrm{OH}){}_3+3[\mathrm{Sn}(\mathrm{OH}){}_4]{}^{2-}\to 2\mathrm{Bi}+3[\mathrm{Sn}(\mathrm{OH}){}_6]{}^{2-}$ describes the stannite test for bismuth. The moles of stannite required per mole of $\mathrm{Bi}(\mathrm{OH}){}_3$ is

From the balanced equation 2 mol of $\mathrm{Bi}(\mathrm{OH}){}_3$ need 3 mol of stannite, so the ratio is $3/2=1.5$. Each $\mathrm{Bi}{}^{3+}$ gains 3 electrons and each $\mathrm{Sn}{}^{2+}$ gives 2, which is the origin of the 2:3 ratio. The black deposit is finely divided bismuth.