Question

Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas

Answer 1

The equation structure for the above mentioned reaction can be written as  

$B_{2} S_{3}+6 H_{2} O \rightarrow 2 H_{3} B O_{3}+3 H_{2} S \uparrow$

Explanation:

Considering the above reaction, When Boron sulfide, reacts with water more violently to form boric acid and hydrogen sulfide gas.

$B_{2} S_{3}+H_{2} O \rightarrow H_{3} B O_{3}+H_{2} S \uparrow$

In order to balance the equation, we can do as follows.There are 2 B - atoms on both sides of the equation, but only 2 H - atoms, and one O - atom on LHS, so we have to balance it by putting 6 in front of water and 2 in front of Boric acid and 3 in front of hydrogen sulphide gas, so that we have 2 B - atoms, 3 - S atoms, 12 H - atoms on both sides of the equation, and it is balanced. Balanced equation is given as,

$B_{2} S_{3}+6 H_{2} O \rightarrow 2 H_{3} B O_{3}+3 H_{2} S \uparrow$

Thus a Balanced equation of the above mentioned reaction is written.