\(a)n_{Zn}=\dfrac{13}{65}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2       0,4              0,2         0,2

\(V_{H_2,đkc}=0,2.24,79=4,958l\\ b)m_{ZnCl_2}=0,2.136=27,2g\\ c)V_{ddHCl}=\dfrac{0,4}{2}=0,2l\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)  

a) \(m_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

Theo PTHH: \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)

b) Theo PTHH: \(n_{MgSO_4}=n_{Mg}=0,15\left(mol\right)\)

\(\Rightarrow m_{MgSO_4}=0,15\cdot120=18\left(g\right)\)

c) Theo PTHH: \(n_{HCl}=n_{Mg}=0,15\left(mol\right)\)

\(V_{dd,H_2SO_4}=\dfrac{0,15}{2,4}=0,0625\left(l\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\\n_{HCl}=2n_{Fe}=0,2\left(mol\right)\end{matrix}\right.\)

a, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

b, \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

c, \(V_{HCl}=\dfrac{0,2}{2,5}=0,08\left(l\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

       0,2        0,3                   0,1         0,3

a) \(m_{muối}=m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)

b) \(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c)\(V_{ddH2SO4}=\dfrac{0,3}{3}=0,1\left(l\right)\)

 Chúc bạn học tốt

2 Al + 3 H2SO4 -----> Al2(SO4)3 + 3 H2
0,2  :     0,3           :        0,1          :   0,3     (mol)
nAl= \(\dfrac{5,4}{27}\approx0,2\) (mol)
a) mAl2(SO4)3 = 0,1.342 = 34,2 (g)
b) VH2 = 0,3.22,4 = 6,72 (lít)
c) VH2SO4 = \(\dfrac{0,3}{3}=0,1\) (lít)

Mong các bạn trả lời

Mình cần gấp lắm

\(a)n_{Zn}=\dfrac{3,9}{65}=0,06mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,06       0,12       0,06      0,06

\(V_{H_2}=0,06.24,79=1,4874l\\ b)V_{HCl}=\dfrac{0,12}{0,8}=0,15l=150ml\)

\(C_M\)\(_{ZnCl_2}=\dfrac{0,06}{0,15}=0,4M\)

\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ b.n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)

0,2        0,4            0,2           0,2

\(V_{H_2}=0,2.24,79=4,958l\\ c.m_{MgCl_2}=0,2.95=19g\\ d.C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}M\)

a, nMg=0,2(mol)

Mg+2HCl=>MgCl2+H2

b, nH2=nMg=0,2(mol)

=>VH2=4,958(l)

c,nMgCl2=nMg=0,2(mol)

=>mMgCl2=19(g)

d,nHCl=2nMg=0,4(mol)

=>cM(HCl)=0,75(M)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2........0,3...........0,1...........0,3\left(mol\right)\\ a.C_{MddH_2SO_4}=\dfrac{0,3}{0,3}=1\left(M\right)\\ b.m_{Al_2\left(SO_4\right)_3}=342.0,1=34,2\left(g\right)\\ c.V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1M\)

b) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,1.342=34,2\left(g\right)\)

c) \(n_{H_2}=n_{H_2SO_4}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)

\(a)n_{Fe}=\dfrac{16,8}{56}=0,3mo\\ n_{H_2SO_4}=\dfrac{196.20}{100.98}=0,4mol\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ \Rightarrow\dfrac{0,3}{1}< \dfrac{0,4}{1}\Rightarrow H_2SO_4.dư\\ n_{Fe}=n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,3mol\\ V_{H_2\left(đkc\right)}=0,3.24,79=7,437l\\ b)C_{\%FeSO_4}=\dfrac{0,2.152}{16,8+196-0,3.2}\cdot100=14,32\%\\ C_{\%H_2SO_4}=\dfrac{\left(0,4-0,3\right).98}{16,8+196-0,3.2}\cdot100=4,62\%\)

\(n_{Al}=\dfrac{2,7}{27}=0,1mol\\ a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1           0,15               0,05               0,15

\(b.V_{H_2}=0,15.24,79=3,7185l\\ c.m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\\ d.C_{M_{H_2SO_4}}=\dfrac{0,15}{0,4}=0,375M\)

D