viết 4 phản ứng hóa học khác nhau để điều chế trực tiếp ra a) dd NaOH b) dd CuCl2 giúp mình vs 14/11/2021 Bởi Jasmine viết 4 phản ứng hóa học khác nhau để điều chế trực tiếp ra a) dd NaOH b) dd CuCl2 giúp mình vs
a, $2Na+2H_2O\to 2NaOH+H_2$ $Na_2O+H_2O\to 2NaOH$ $Na_2O_2+H_2O\to 2NaOH+\dfrac{1}{2}O_2$ $Ba(OH)_2+Na_2SO_4\to BaSO_4+2NaOH$ b, $Cu+Cl_2\buildrel{{t^o}}\over\to CuCl_2$ $CuO+2HCl\to CuCl_2+H_2O$ $Cu(OH)_2+2HCl\to CuCl_2+2H_2O$ $Cu+2HCl+\dfrac{1}{2}O_2\to CuCl_2+H_2O$ Bình luận
\(\begin{array}{l}a)\\2Na + 2{H_2}O \to 2NaOH + {H_2}\\N{a_2}O + {H_2}O \to 2NaOH\\2NaCl + 2{H_2}O \to 2NaOH + C{l_2} + {H_2}\\Ba{(OH)_2} + N{a_2}S{O_4} \to 2NaOH + BaS{O_4}\\b)\\Cu + C{l_2} \to CuC{l_2}\\Cu + 2FeC{l_3} \to 2FeC{l_2} + CuC{l_2}\\CuO + 2HCl \to CuC{l_2} + {H_2}O\\CuS + 2HCl \to CuC{l_2} + {H_2}S\end{array}\) Bình luận
a,
$2Na+2H_2O\to 2NaOH+H_2$
$Na_2O+H_2O\to 2NaOH$
$Na_2O_2+H_2O\to 2NaOH+\dfrac{1}{2}O_2$
$Ba(OH)_2+Na_2SO_4\to BaSO_4+2NaOH$
b,
$Cu+Cl_2\buildrel{{t^o}}\over\to CuCl_2$
$CuO+2HCl\to CuCl_2+H_2O$
$Cu(OH)_2+2HCl\to CuCl_2+2H_2O$
$Cu+2HCl+\dfrac{1}{2}O_2\to CuCl_2+H_2O$
\(\begin{array}{l}
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
N{a_2}O + {H_2}O \to 2NaOH\\
2NaCl + 2{H_2}O \to 2NaOH + C{l_2} + {H_2}\\
Ba{(OH)_2} + N{a_2}S{O_4} \to 2NaOH + BaS{O_4}\\
b)\\
Cu + C{l_2} \to CuC{l_2}\\
Cu + 2FeC{l_3} \to 2FeC{l_2} + CuC{l_2}\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
CuS + 2HCl \to CuC{l_2} + {H_2}S
\end{array}\)