Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 113: 3.16b

Answer

$B_2S_3(s) + 6H_2O(l)\rightarrow 2H_3BO_3(aq)+ 3H_2S(g)$

Work Step by Step

Boron sulfide is $B_2S_3(s)$, water is $H_2O$, boric acid is $H_3BO_3$ and hydrogen sulfide gas is $H_2S$. $B_2S_3(s) + H_2O(l)\rightarrow H_3BO_3(aq)+ H_2S(g)$ Add a coefficient of 2 to $H_3BO_3$ to balance $B$. Add a coefficient of 3 to $H_2S$ to balance $S$. $B_2S_3(s) + H_2O(l)\rightarrow 2H_3BO_3(aq)+ 3H_2S(g)$ Add a coefficient of 6 to $H_2O$ to balane $H$ and $O$. $B_2S_3(s) + 6H_2O(l)\rightarrow 2H_3BO_3(aq)+ 3H_2S(g)$
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