4
a
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
0,05------>0,1-------->0,05
b
\(n_{Ba\left(OH\right)_2}=0,05.1=0,05\left(mol\right)\)
\(CM_{BaCl_2}=\dfrac{0,05}{0,05+0,08}=\dfrac{5}{13}M\)
\(a=CM_{HCl}=\dfrac{0,1}{0,08}=1,25M\)
5
\(NaOH+HCl\rightarrow NaCl+H_2O\)
x---------->x------>x
\(KOH+HCl\rightarrow KCl+H_2O\)
2x------>2x------>2x
\(n_{HCl}=x+2x=0,15.2=0,3\\ \Rightarrow x=0,1\)
a
\(CM_{NaOH}=\dfrac{0,1}{0,1}=1M\\ CM_{KOH}=\dfrac{0,2}{0,1}=2M\)
b
\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\\m_{KCl}=0,2.74,5=14,9\left(g\right) \)
6
\(n_{H_2SO_4}=0,06.1=0,06\left(mol\right)\)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,1-------->0,05----->0,05
Xét \(\dfrac{0,1}{2}< \dfrac{0,06}{1}\Rightarrow n_{H_2SO_4dư}=0,06-0,05=0,01\left(mol\right)\)
a
\(CM_{Na_2SO_4}=\dfrac{0,05}{0,06}=\dfrac{5}{6}M\)
\(CM_{H_2SO_4}=\dfrac{0,01}{0,06}=\dfrac{1}{6}M\)
b
Nhúng giấy quỳ tím vào dung dịch A, quỳ chuyển sang màu đỏ vì trong A ngoài muối còn có `H_2SO_4` dư.
7
\(n_{KOH}=0,12.1=0,12\left(mol\right)\)
\(n_{HNO_3}=0,08.0,5=0,04\left(mol\right)\\ n_{HCl}=0,08.0,5=0,04\left(mol\right)\)
\(KOH+HCl\rightarrow KCl+H_2O\)
0,04<---0,04--->0,04
\(n_{KOH.dư}=0,12-,0,04=0,08\left(mol\right)\)
\(KOH+HNO_3\rightarrow KNO_3+H_2O\)
0,04<---0,04----->0,04
\(n_{KOH.dư}=0,08-0,04=0,04\left(mol\right)\)
\(CM_{KCl}=\dfrac{0,04}{0,12+0,08}=0,2M\\ CM_{KNO_3}=\dfrac{0,04}{0,12+0,08}=0,2M\\ CM_{KOH}=\dfrac{0,04}{0,12+0,08}=0,2M\)
8
\(n_{HNO_3}=0,4.1=0,4\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
2x------>2x---------->2x
\(Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\)
x------------>2x---------->x
\(n_{HNO_3}=2x+2x=0,4\\ \Rightarrow x=0,1\)
a
\(CM_{NaOH}=\dfrac{2.0,1}{0,2}=1M\)
\(CM_{Ba\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5M\)
b
\(m_{NaNO_3}=0,2.85=17\left(g\right)\\ m_{Ba\left(NO_3\right)_2}=0,1.261=26,1\left(g\right)\)
3
\(n_{HNO_3}=\dfrac{50.12,6\%}{100\%}:63=0,1\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
0,1<------0,1------->0,1
\(m_{NaNO_3}=0,1.85=8,5\left(g\right)\)
\(CM_{NaOH}=\dfrac{0,1}{0,2}=0,5M\)
