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\(n_{Ca}=\dfrac{12}{40}=0,3\left(mol\right)\)
PT: \(Ca+H_2SO_4\rightarrow CaSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{CaSO_4}=n_{H_2}=n_{Ca}=0,3\left(mol\right)\)
a, \(V_{H_2}=0,3.24,79=7,437\left(l\right)\)
b, \(m_{CaSO_4}=0,3.136=40,8\left(g\right)\)
c, \(V_{H_2SO_4}=\dfrac{0,3}{1,6}=0,1875\left(l\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
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)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) PTHH : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow C_{MddHCl}=\dfrac{0,4}{0,2}=2M\)
c) \(n_{H2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H2\left(dkc\right)}=0,2.24,79=4,958\left(l\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2........0,3...........0,1...........0,3\left(mol\right)\\ a.C_{MddH_2SO_4}=\dfrac{0,3}{0,3}=1\left(M\right)\\ b.m_{Al_2\left(SO_4\right)_3}=342.0,1=34,2\left(g\right)\\ c.V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)
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)
\(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
_____0,4<---0,4<--------0,4<----0,4
=> mZn = 0,4.65 = 26 (g)
=> \(\left\{{}\begin{matrix}\%Zn=\dfrac{26}{51,6}.100\%=50,388\%\\\%Cu=\dfrac{51,6-26}{51,6}.100\%=49,612\text{%}\end{matrix}\right.\)
b)
mZnSO4 = 0,4.161 = 64,4 (g)
c)
\(V_{ddH_2SO_4}=\dfrac{0,4}{2}=0,2\left(l\right)\)
\(a)n_{Zn}=\dfrac{13}{65}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2 0,2
\(V_{H_2,đkc}=0,2.24,79=4,958l\\ b)m_{ZnCl_2}=0,2.136=27,2g\\ c)V_{ddHCl}=\dfrac{0,4}{2}=0,2l\)