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\(a,m_{HCl}=0,5.36,5=18,25(g)\\ b,Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=0,25(mol)\\ \Rightarrow V_{H_2}=0,25.22,4=5,6(l)\\ m_{Zn}=0,25.65=16,25(g)\)
bn có thẻ trình bay và ghi rõ các bước ra được khum cô mik bảo cách trình bày sai
a) PTHH: Zn + H2SO4 -> ZnSO4 + H2
nH2= 0,15(mol)
=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)
b) mZn=0,15.65=9,75(g)
c) CMddH2SO4= 0,15/ 0,05=3(M)
d) mZnSO4= 161. 0,15=24,15(g)
\(PTHH:Zn+2HCl->ZnCl_2+H_2\)
ap dung DLBTKL ta co
\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)
\(=>m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}\\ =>m_{H_2}=13+14,6-27,2\\ =>m_{H_2}=0,4\left(g\right)\)
nZn = 52 : 65 = 0,8 (mol)
pthh : Zn + 2HCl ---> ZnCl2 +H2
0,8--->1,6----------------->0,8 (mol)
=> mHCl = 1,6 . 36,5 = 58,4 (g)
VH2 = 0,8 . 22,4 = 17,92 (l)
nFe3O4 = 9,28 : 232 = 0,04 (mol )
pthh : Fe3O4 + 4H2 -t--> 3Fe + 4H2O
LTL :
0,04/1 < 0,8/4 => H2 DU
theo pthh , nFe = 3nFe3O4 = 0,12 (mol)
=> m Fe = 0,12 . 56= 6,72 (g)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)
b, mZn = 0,15.65 = 9,75 (g)
c, CM (H2SO4) = 0,15/0,05 = 3 M
d, mZnSO4 = 0,15.161 = 24,15 (g)
Bạn tham khảo nhé!
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + H2SO4 ---> ZnSO4 + H2
0,2--->0,2--------->0,2------>0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)
`Zn+H_2SO_4->ZnSO_4+H_2`(to)
0,45-------------------0,45------0,45mol
`n_(Zn)=(29,25)/65=0,45mol`
`m_(ZnSO_4)=0,45.161=72,45g`
`V_(H_2)=0,45.22,4=10,08l`
c) `H_2+CuO->Cu+H_2O`(to)
0,45--------0,45 mol
`n_(Cu)=40/80=0,5 mol`
=>Cu dư , 0,05 mol
`m_(chất rắn)=0,45.64+0,05.80=32,8g`
\(n_{Zn}=\dfrac{m}{M}=\dfrac{29,25}{65}=0,45\left(mol\right)\)
a) \(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1 1 1 1
0,45 0,45 0,45 0,45
b) \(m_{ZnSO_4}=n.M=0,45.\left(65+32+16.4\right)=51,03\left(g\right)\\ V_{H_2}=n.24,79=0,45.24,79=11,1555\left(l\right)\)
c) \(n_{CuO}=\dfrac{m}{M}=\dfrac{40}{\left(64+16\right)}=0,5\left(mol\right)\)
\(PTHH:CuO+H_2\rightarrow Cu+H_2O\)
1 1 1 1
0,5 0,5 0,5 0,5
\(m_{Cu}=0,5.64=32\left(g\right).\)
a) Có: \(n_{HCl}=0,5mol\)
\(\Rightarrow m_{HCl}=0,5.\left(1+35,5\right)=18,25g\)
b)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Có: \(n_{HCl}=0,5mol\)
Theo PTHH:
\(\Rightarrow n_{H_2}=0,25mol\)
\(\Rightarrow V_{H_2}=0,25.22,4=5,6l\)
\(n_{Zn}=0,25mol\)
\(\Rightarrow m_{Zn}=0,25.65=16,25g\)