Answer
a. $ KHCO_3(s) + HBr(aq)\longrightarrow CO_2(g) + H_2O(l) + KBr(aq)$
b. $Ca(s) + H_2S{O_4}(aq) \longrightarrow H_2(g) + CaSO_4(aq)$
c. $H_2SO_4(aq) + Ca(OH)_2(s) \longrightarrow CaSO_4(aq) + 2H_2O(l)$
d. $Na_2CO_3(s) + H_2SO_4(aq) \longrightarrow CO_2(g) + H_2O(l) + Na_2SO_4(aq)$
Work Step by Step
a. This is a reaction between an acid and a bicarbonate. Thus, we know that it will produce $CO_2(g)$, $H_2O(l)$, using the bicarbonate and a proton from the acid, and a salt with the remaining ions.
Remaining ions: $K^+$ and $Br^-$
$ KHCO_3(s) + HBr(aq)\longrightarrow CO_2(g) + H_2O(l) + KBr(aq)$
b. This is a reaction between an acid and a metal. Therefore, we know that it will produce $H_2(g)$ and a salt made with the metal ion and the conjugate base of the acid.
- Metal ion: $Ca^{2+}$
- Conjugate base: $S{O_4}^{2-}$
Resulting salt: $CaSO_4$
$Ca(s) + H_2S{O_4}(aq) \longrightarrow H_2(g) + CaSO_4(aq)$
c. This is a reaction between an acid and a hydroxide. Therefore, we know that it will produce $H_2O(l)$, with the protons from the acid, and the hydroxide ions. The remaining ions will produce a salt.
Remaining ions: $S{O_4}^{2-}$ and $Ca^{2+}$
$H_2SO_4(aq) + Ca(OH)_2(s) \longrightarrow CaSO_4(aq) + H_2O(l)$
- Balance the equation:
$H_2SO_4(aq) + Ca(OH)_2(s) \longrightarrow CaSO_4(aq) + 2H_2O(l)$
d. This is a reaction between an acid and a carbonate. Firstly, the Carbonate ion will react with 2 protons from the acid, producing $H_2CO_3$, which will break down into $H_2O(l)$ and $CO_2(g)$. The remaining ions: $Na^{+}$ and $S{O_4}^{2-}$ will produce a salt: $Na_2SO_4$.
$Na_2CO_3(s) + H_2SO_4(aq) \longrightarrow CO_2(g) + H_2O(l) + Na_2SO_4(aq)$
