Bài 21:
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\\ PTHH:MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
0,1---->0,1--------->0,1
a. \(V_{dd.H_2SO_4}=\dfrac{0,1}{1}=0,1\left(l\right)=100\left(ml\right)\)
b. \(CM_{MgSO_4}=\dfrac{0,1}{0,1}=1M\)
Bài 22:
\(n_{CuO}=\dfrac{16}{160}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{100.19,6\%}{100\%}:98=0,2\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
0,1----->0,1------>0,1
Xét \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow n_{H_2SO_4.dư}=0,2-0,1=0,1\left(mol\right)\)
a. Tìm V cái gì vậy?
b. \(C\%_{CuSO_4}=\dfrac{0,1.160.100\%}{16+100}=13,79\%\)
\(C\%_{H_2SO_4}=\dfrac{0,1.98.100\%}{16+100}=8,45\%\)
B23:
`n_Na=\frac{3,45}{23}=0,15(mol)`
PTHH:
`2Na+2H_2O\rightarrow2NaOH+H_2`
0,15---------------->0,15----->0,075
a. \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
b. \(C\%_{NaOH}=\dfrac{40.0,15.100\%}{3,45+m_{H_2O}-m_{H_2}}\) (cái này thiếu dữ kiện m gam nước nhé=)
\(CM_{NaOH}=\dfrac{0,15}{0,05}=3\left(M\right)\)
B24:
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,25------------->0,5
a. \(CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b.
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,5--------->0,25----->0,25
\(m_{dd.H_2SO_4.cần}=\dfrac{0,25.98.100\%}{20\%}=122,5\left(g\right)\\ V_{dd.H_2SO_4.cần}=\dfrac{122,5}{1,14}=107,46\left(ml\right)=0,10746\left(l\right)\)
c. \(CM_{Na_2SO_4}=\dfrac{0,25}{0,5+0,10746}=0,4\left(M\right)\)
$HaNa$♬
