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1:
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
______0,2------>0,2------------------->0,2_____(mol)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b) \(V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)\)
2:
a)
\(n_{HCl}=2.0,2=0,4\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
______0,2<------0,4------------------>0,2______(mol)
=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)
b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
a, nHCl = 0,27 ( mol )
=> n(H) = 0,27 ( mol )
=> nH2 = 0,135 ( mol )
=> VH2 = 3,024 ( l )
b, Có n(Cl) = nHCl = 0,27 ( mol )
=> m(Cl) = 9,585 ( g )
=> mMuối = mKl + m(Cl) = 3,88 + 9,585 = 13,465 ( g )
nHCl = 0,135.2 = 0,27 mol
Phản ứng vừa đủ , áp dụng ĐLBT nguyên tố H => nH2 = \(\dfrac{0,27}{2}\)=0,135mol
=> V H2 = 0,135. 22,4 = 3,024 lít
b) Hỗn hợp kim loại + HCl --> Hỗn hợp muối + H2
Áp dụng ĐLBT khối lượng => m hh kim loại + m HCl = m hỗn hợp muối + mH2
=> m muối = 3,88 + 0,27.36,5 - 0,135.2 =13,465 gam
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,1-->0,2------>0,1-->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
mZnCl2 = 0,1.136 = 13,6 (g)
b) \(C\%_{dd.HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)
c) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,15}{1}\) => H2 hết, O2 dư
PTHH: 2H2 + O2 --to--> 2H2O
0,1--------------->0,1
=> mH2O = 0,1.18 = 1,8 (g)
a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
____0,1_____0,2______0,1_____0,1 (mol)
\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
b, \(C_{M_{HCl}}=\dfrac{0,2}{0,1}=1\left(M\right)\)
c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.
Theo PT: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\)
⇒ m chất rắn = mCuO (dư) + mCu = 0,05.80 + 0,1.64 = 10,4 (g)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Cu}=20-5,6=14,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\\%m_{Cu}=72\%\end{matrix}\right.\)
\(a) Fe + 2HCl \to FeCl_2\\ b) n_{HCl} = \dfrac{182,5.5\%}{36,5} = 0,25(mol)\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,125(mol)\\ \Rightarrow m_{Fe} = 0,125.56 = 7(gam) ; V = 0,125.22,4 = 2,8(lít)\\ c) m_{dd\ sau\ phản\ ứng} = m_{Fe} + m_{dd\ HCl} - m_{H_2} = 7 + 182,5 - 0,125.2 = 189,25(gam)\\ C\%_{FeCl_2} = \dfrac{0,125.127}{189,25}.100\% = 8,39\%\)
\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05
b) \(n_{H2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddHCl}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
Chúc bạn học tốt
\(a.PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ n_{H_2}=0,2.2=0,4\left(mol\right)\\ V_{H_2}=0,4.22,4=8,96\left(l\right)\\ c.n_{HCl}=n_{Zn}=0,2mol\\ C_{MHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(c,C_M=\dfrac{n}{V}=\dfrac{0,4}{0,1}=4M\)
\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2-->0,4----------------->0,2
\(maxV_{H_2}=0,2.22,4=4,48\left(l\right)\\ b,C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2M\\ c,n_{Cl}=m_{HCl}=0,4\left(mol\right)\\ BTKL:m_{muối}=m_{KL}+m_{Cl}=5,5+0,4.35,5=19,7\left(g\right)\)