$BaCl_2 + KHSO_4 \rightarrow BaSO_4 \downarrow + HCl + KCl$

$Ba^{2+} + HSO_4^- \rightarrow BaSO_4 \downarrow + H^+$

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$NaHCO_3 + CaCl_2 \rightarrow CaCO_3 \downarrow + HCl + NaCl$

$HCO_3^- + Ca^{2+} \rightarrow CaCO_3 \downarrow + H^+$

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$2KHSO_4 + Ba(HCO_3)_2 \rightarrow BaSO_4 \downarrow + 2CO_2 \uparrow + K_2SO_4 + 2H_2O$

$Ba^{2+} + HSO_4^- + HCO_3^- \rightarrow BaSO_4 \downarrow + CO_2 \uparrow + H_2O$

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$H_2S + 2FeCl_3 \rightarrow S \downarrow + 2FeCl_2 + 2H_2O$

$H_2S + 2Fe^{3+} \rightarrow S \downarrow + 2Fe^{2+} + 2H^+$

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$3Na_2CO_3 + 2FeCl_3 \rightarrow Fe_2(CO_3)_3 \downarrow + 6NaCl$

$3CO_3^{2-} + 2Fe^{3+} \rightarrow Fe_2(CO_3)_3 \downarrow$

$\text{#Bbi H ><}$

Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
a,BaC{l_2} + KHS{O_4}\\
B{a^{2 + }} + 2C{l^ - } + {K^ + } + HS{O_4}^ -  \to BaS{O_4} + 2C{l^ - } + {K^ + } + {H^ + }\\
B{a^{2 + }} + S{O_4}^{2 - } \to BaS{O_4}\\
b,NaHC{O_3} + CaC{l_2}\\
N{a^ + } + HC{O_3}^ -  + C{a^{2 + }} + 2C{l^ - } \to CaC{O_3} + N{a^ + } + C{l^ - } + C{O_2} + {H_2}O\\
C{O_3}^{2 - } + C{a^{2 + }} \to CaC{O_3}\\
2{H^ + } + C{O_3}^{2 - } \to C{O_2} + {H_2}O\\
c,KHS{O_4} + Ba{(HC{O_3})_2}\\
{K^ + } + HS{O_4}^ -  + B{a^{2 + }} + 2HC{O_3}^ -  \to BaS{O_4} + C{O_2} + {H_2}O + 2{K^ + } + S{O_4}^{2 - }\\
B{a^{2 + }} + S{O_4}^{2 - } \to BaS{O_4}\\
2{H^ + } + C{O_3}^{2 - } \to C{O_2} + {H_2}O\\
d,{H_2}S + FeC{l_3}\\
{H_2}S + F{e^{3 + }} + 3C{l^ - } \to F{e^{2 + }} + {H^ + } + 3C{l^ - } + S\\
{H_2}S + 2F{e^{3 + }} \to 2F{e^{2 + }} + 2{H^ + } + S\\
e,N{a_2}C{O_3} + FeC{l_3} + {H_2}O\\
6N{a^ + } + 3C{O_3}^{2 - } + 2F{e^{3 + }} + 6C{l^ - } + 3{H_2}O \to 2Fe{(OH)_3} + 3C{O_2} + 6N{a^ + } + 6C{l^ - }\\
2F{e^{3 + }} + 3C{O_3}^{2 - } + 3{H_2}O \to 2Fe{(OH)_3} + 3C{O_2}
\end{array}\)