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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a/2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\)
\(b/30ml=0,03l\\ n_{H_2SO_4}=0,5.0,03=0,0015\left(mol\right)\\ n_{Al}=\dfrac{0,0015.2}{3}=0,001\left(mol\right)\\ m_{Al}=0,001.27=0,027\left(g\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,0015}{2}=0,00075\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=0,00075.342=0,2565\left(g\right)\)
\(c/n_{H_2}=\dfrac{0,0015.3}{3}=0,0015\left(mol\right)\\ V_{H_2}=0,0015.24,79=0,037185\left(l\right)\)
\(a.2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\\ b.n_{Al}=1,5.0,5.0,03=0,0375mol\\ m_{Al}=0,0375.27=1,0125g\\ m_{Al_2\left(SO_4\right)_3}=342\cdot\dfrac{1}{3}\cdot0,03\cdot0,5=1,71g\\V_{H_2}=24,79.0,5.0,03=0,37185L\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(n_{HCl\left(bđ\right)}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,4<--0,8<----0,4<----0,4
=> mHCl(dư) = (1-0,8).36,5 = 7,3 (g)
c) mFe = 0,4.56 = 22,4 (g)
mFeCl2 = 0,4.127 = 50,8 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{H_2SO_4}=\dfrac{200\cdot9,8\%}{100\%}=19,6\left(g\right)\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\\ PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ \Rightarrow n_{KOH}=2n_{H_2SO_4}=0,4\left(mol\right)\\ \Rightarrow m_{KOH}=0,4\cdot\left(39+16+1\right)=22,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2Al+3H2SO4->Al2(SO4)3+3H2
0,1----------------------0,075----0,15
n H2=0,15 mol
=>mAl=0,1.27=2,7g
=>m Al2(SO4)3=0,075.342=25,65g
a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
b, Số mol \(H_2SO_4\) là: \(n_1=V.C_M=0,5.0,5=0,25\) (mol)
Số mol \(Na_2SO_4\) là \(n_2=\dfrac{28,4}{142}=0,2\) (mol)
Do \(n_2< n_1\) nên \(H_2SO_4\) còn dư
Suy ra số mol \(Na_2O\) tham gia phản ứng là: \(n=n_2=0,2\) (mol)
Khối lượng là: \(m_{Na_2O}=0,2.62=12,4g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(m_{HCl}=\dfrac{200.10,95}{100}=21,9\left(g\right)\)
=> \(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
b) \(n_{CaCO_3}=\dfrac{a}{100}=0,01a\left(g\right)\)
\(n_{NaOH}=0,05.2=0,1\left(mol\right)\)
PTHH: CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
______0,01a---->0,02a---->0,01a->0,01a___________(mol)
NaOH + HCl --> NaCl + H2O
_0,1----->0,1___________________________________(mol)
=> 0,02a = 0,6 - 0,1
=> a = 25 (g)
c) \(V_{CO_2}=0,01.25.22,4=5,6\left(l\right)\)
d) \(\left\{{}\begin{matrix}C\%\left(CaCl_2\right)=\dfrac{0,25.111}{25+200-0,25.44}.100\%=12,97\%\\C\%\left(HCl_{dư}\right)=\dfrac{0,1.36,5}{25+200-0,25.44}.100\%=1,705\%\end{matrix}\right.\)
\(SO_3+2NaOH\rightarrow Na_2SO_4+H_2O\\ n_{SO_3}=\dfrac{8}{40}=0,2\left(mol\right)\Rightarrow n_{Na_2SO_4}=n_{SO_3}=0,2\left(mol\right);n_{NaOH}=2.0,2=0,4\left(mol\right)\\ m_{NaOH}=0,4.40=16\left(g\right);m_{Na_2SO_4}=142.0,2=28,4\left(g\right)\)