K
kobato_2509
$2NaOH + CuSO_4 \rightarrow Na_2SO_4 + Cu(OH)_2$
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$Fe + H_2SO_4 \rightarrow FeSO_4 + H_2$
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$o_2+n_2------->no$
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$2KNO_3 \xrightarrow{t^o} 2KNO_2 + O_2$
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$H_2O + NH_3 \to NH_4^+ + OH^-$$Ba(OH)_2+H_2SO_4---------->BaSO_4+H_2O$
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