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a. \(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{12}{24}=0,5\left(mol\right)\)
- Mol theo PTHH : \(1:2:1:1\)
- Mol theo phản ứng : \(0,5\rightarrow1\rightarrow0,5\rightarrow0,5\)
\(\Rightarrow n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)
b. Từ a. \(\Rightarrow n_{HCl}=1\left(mol\right)\)
\(\Rightarrow m_{HCl}=n_{HCl}.M_{HCl}=1.\left(1+35,5\right)=36,5\left(g\right)\)
c. \(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\)
- Mol theo PTHH : \(2:1:2\)
- Mol theo phản ứng : \(0,6\leftarrow0,3\rightarrow0,6\)
\(\Rightarrow n_{H_2O}=0,6\left(mol\right)\)
\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,6.\left(2+16\right)=10,8\left(g\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{1}{15}.56=\dfrac{56}{15}\left(g\right)\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Ta có: \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{HCl}=2.0,2=0,4\left(mol\right);n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,m_{HCl}=0,4.36,5=14,6\left(g\right)\\ c,n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Vì:\dfrac{0,2}{1}< \dfrac{0,3}{1}\Rightarrow CuOdư\\ n_{Cu}=n_{H_2}=0,2\left(mol\right)\\ m_{Cu}=0,2.64=12,8\left(g\right)\)
\(n_{Mg}=\dfrac{13}{24}=0,54mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,54 0,54 ( mol )
\(m_{MgCl_2}=0,54.95=51,3g\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,54 0,54 ( mol )
\(m_{Cu}=0,54.64=34,56g\)
được tạo thành.
a. PTHH: 2Al + 6HCl \(\rightarrow\) 2 AlCl3 + 3H2
b. Ta có: nAl = \(\frac{6,75}{27}\) = 0,25 mol
Theo p.trình: nH2 = \(\frac{3}{2}\)nAl = \(\frac{3}{2}\). 0,25= 0,375 mol
\(\Rightarrow\) VH2 = 0,375. 22,4 = 8,4 (lít).
c. Theo p.trình: nHCl = 3.nAl = 3.0,5= 0,75 mol
\(\Rightarrow\) mHCl = 0,75. 36,5 = 27,375g
d. Theo p.trình: nAlCl3 = nAl = 0,25 mol
\(\Rightarrow\) mAlCl3 = 0,25.133,5= 33,375g
a) pthh: 3Al+6H->2AlCl3+3H2. b) nAl=6,75/27=0,3 mol ->nH2= 3/2nAl=0,5 -> vH2=11.2 l. c) ta co nHCl=3nAl=0,9mol -> kl HCl pư =0,9×36,5=32,9g
d) nAlCl3=nAl=0,3 mol
->ko Alcl3=0.3×(27+35,5×3)=40.1g
a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1--->0,2------->0,1---->0,1
=> mHCl = 0,2.36,5 = 7,3(g)
b) mMgCl2 = 0,1.95 = 9,5 (g)
c) VH2 = 0,1.22,4 = 2,24(l)
a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1--->0,2------->0,1---->0,1
=> mHCl = 0,2.36,5 = 7,3(g)
b) mMgCl2 = 0,1.95 = 9,5 (g)
c) VH2 = 0,1.22,4 = 2,24(l)
a.b.\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15 0,3 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{HCl}=0,3.36,5=10,95g\)
c.\(n_{H_2}=0,15.60\%=0,09mol\)
\(Ag_2O+H_2\rightarrow\left(t^o\right)2Ag+H_2O\)
0,09 0,18 ( mol )
\(m_{Ag}=0,18.108=19,44g\)
\(a.n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ TheoPT:n_{H_2}=n_{Mg}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\\ b.TheoPT:n_{HCl}=2n_{Mg}=0,3\left(mol\right)\\ \Rightarrow m_{Mg}=0,3.36,5=10,95\left(g\right)\\ c.n_{H_2\left(pứ\right)}=0,15.60\%=0,054\left(g\right)\\ H_2+Ag_2O-^{t^o}\rightarrow2Ag+H_2O\\ n_{Ag}=2n_{H_2}=0,108\left(mol\right)\\ \Rightarrow m_{Ag}=0,108.108=11,664\left(g\right)\)