Bài 9:
a)
\(m_{H_2SO_4}=200.29,4\%=58,8\left(g\right)\\ n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
1,2<-----0,6
b)
\(m_{KOH}=1,2.56=84\left(g\right)\\ \rightarrow m_{dd.KOH}=\dfrac{84}{20\%}=420\left(g\right)\)
c)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
1,2<--------0,6
\(m_{NaOH}=1,2.40=48\left(g\right)\\ \rightarrow m_{dd.NaOH}=\dfrac{48}{20\%}=240\left(g\right)\\ \rightarrow V_{dd.NaOH}=\dfrac{240}{1,045}\approx229,67\left(ml\right)\)
Bài 10:
a)
\(n_{H_2SO_4}=0,5.1=0,5\left(mol\right)\)
PTHH: \(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)
0,5<----------0,5
b)
\(m_{Ca\left(OH\right)_2}=0,5.74=37\left(g\right)\\ \rightarrow m_{dd.Ca\left(OH\right)_2}=\dfrac{37}{20\%}=185\left(g\right)\)
c)
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
1<--------0,5
\(\rightarrow m_{KOH}=1.56=56\left(g\right)\\ \rightarrow m_{dd.KOH}=\dfrac{56}{5,6\%}=1000\left(g\right)\\ \rightarrow V_{dd,KOH}=\dfrac{1000}{1,045}=956,94\left(ml\right)\)
Bài 11:
a)
\(n_{H_2SO_4}=2.0,2=0,4\left(mol\right)\)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,8<------0,4
b)
\(m_{NaOH}=0,8.40=32\left(g\right)\\ \rightarrow m_{dd.NaOH}=\dfrac{32}{10\%}=320\left(g\right)\)
c)
PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)
0,4<--------0,4
\(m_{Ba\left(OH\right)_2}=0,4.171=68,4\left(g\right)\\ \rightarrow m_{dd.Ba\left(OH\right)_2}=\dfrac{68,4}{17,1\%}=400\left(g\right)\\ \rightarrow D_{dd.Ba\left(OH\right)_2}=\dfrac{400}{200}=2\left(g/ml\right)\)
Bài 12:
a)
Gọi \(\left\{{}\begin{matrix}n_{Na}=a\left(mol\right)\\n_K=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH:
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
a------------------->a-------->0,5a
\(2K+2H_2O\rightarrow2KOH+H_2\)
b------------------->b-------->0,5b
\(NaOH+HCl\rightarrow NaCl+H_2O\)
a---------->a------->a
\(KOH+HCl\rightarrow KCl+H_2O\)
b---------->b------->b
=> \(\left\{{}\begin{matrix}0,5a+0,5b=0,1\\58,5a+74,5b=13,3\end{matrix}\right.\)
=> a = b = 0,1
=> \(\left\{{}\begin{matrix}m_{Na}=0,1.23=2,3\left(g\right)\\m_K=0,1.39=3,9\left(g\right)\end{matrix}\right.\)
b)
\(n_{HCl}=0,1+0,1=0,2\left(mol\right)\\ \rightarrow V_{dd.HCl}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
c)
PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,1<---0,1
=> mCuO = 0,1.80 = 8 (g)
