19
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ n_{H_2SO_4}=0,06.1=0,06\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,1------->0,05------->0,05
Xét \(\dfrac{0,1}{2}< \dfrac{0,06}{1}\Rightarrow H_2SO_4\) dư
a. A gồm \(\left\{{}\begin{matrix}n_{H_2SO_4dư}=0,06-0,05=0,01\left(mol\right)\\n_{Na_2SO_4}=0,05\left(mol\right)\end{matrix}\right.\)
Trong A: \(\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,01}{0,06}=\dfrac{1}{6}M\\CM_{Na_2SO_4}=\dfrac{0,05}{0,06}=\dfrac{5}{6}M\end{matrix}\right.\)
b. Nhúng giấy quỳ tím vào dung dịch (A) quỳ chuyển sang màu đỏ vì axit `H_2SO_4` dư
20
\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,1--------------->0,2
a.
\(a=CM_{NaOH}=\dfrac{0,2}{0,2}=1M\)
b.
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,2------->0,1
\(V_{H_2SO_4}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)
21
\(n_{CuCl_2}=\dfrac{26}{135}\left(mol\right)\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
\(\dfrac{26}{135}\)------------------>\(\dfrac{26}{135}\)
a
\(m=m_{Cu\left(OH\right)_2}=\dfrac{26}{135}.98\approx18,874\left(g\right)\)
b
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(\dfrac{26}{135}\)------->\(\dfrac{26}{135}\)
\(m_{chất.rắn}=m_{CuO}=\dfrac{26}{135}.80\approx15,407\left(g\right)\)
Câu 19
\(a)n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right);n_{H_2SO_4}=0,06.1=0,06\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ \dfrac{0,1}{2}< \dfrac{0,06}{1}\Rightarrow H_2SO_4.dư\\ n_{Na_2SO_4}=0,1:2=0,05\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,06-0,1:2=0,01\left(mol\right)\\ C_M_{Na_2SO_4}=\dfrac{0,05}{0,06}=\dfrac{5}{6}\left(M\right)\\ C_{MH_2SO_4\left(dư\right)}=\dfrac{0,01}{0,06}=\dfrac{1}{6}\left(M\right)\)
b) Vì H2SO4 dư nên nhúng giấy vào dd (A) thì quỳ tím chuyển sang màu đỏ.
Câu 20
\(a)n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\\ Na_2O+H_2O\xrightarrow[]{}2NaOH\\ n_{NaOH}=0,1.2=0,2\left(mol\right)\\ C_{MNaOH}=\dfrac{0,2}{0,2}=1\left(M\right)\\ b)2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ V_{H_2SO_4}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)
Câu 21
\(a)n_{CuCl_2}=\dfrac{26}{135}\approx0,19\left(mol\right)\\ CuCl_2+2NaOH\xrightarrow[]{}Cu\left(OH\right)_2+2NaCl\\ n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,19mol\\ m_{Cu\left(OH\right)_2}=0,19.98=18,62\left(g\right)\\ b)Cu\left(OH\right)_2\xrightarrow[]{t^0}CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=0,19mol\\ m_{CuO}=0,19.80=15,2\left(g\right)\)
