Cho 1,95 gam kẽm tác dụng với dung dịch có hòa tan 1,47 gam H2SO4 nguyên chất.
a. Viết phương trình hóa học.
b. Tính khối lượng chất còn dư sau phản ứng.
c. Tính thể tích khí hidro (đktc) tạo thành sau phản ứng.
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a)\(PTHH:Zn+H_2SO_4\underrightarrow{ }ZnSO_4+H_2\)
b)\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(m\right)\);\(n_{H_2SO_4}=\dfrac{1,57}{98}=0,16\left(m\right)\)
\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)
ta có tỉ lệ:\(\dfrac{0,3}{1}>\dfrac{0,16}{1}->Zndư\)
\(n_{Zn\left(dư\right)}=0,3-0,16=0,14\left(m\right)\)
\(m_{Zn\left(dư\right)}=0,14.65=9,1\left(g\right)\)
c)\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)
tỉ lệ :1 1 1 1
số mol :0,16 0,16 0,16 0,16
\(V_{H_2}=0,16.22,4=3,584\left(l\right)\)
\(n_{Zn}=\dfrac{m}{M}=\dfrac{1,95}{65}=0,03\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{1,57}{98}\approx0,016\left(mol\right)\)
\(PT:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(\dfrac{n_{Zn\left(ĐB\right)}}{n_{Zn\left(PT\right)}}=\dfrac{0,03}{1}>\dfrac{n_{H_2SO_4\left(ĐB\right)}}{n_{H_2SO_4}\left(PT\right)}=\dfrac{0,016}{1}\)
\(\Rightarrow\) Zn dư , H2SO4 hết , tính theo H2SO4
b, Theo PT : \(n_{zn}=n_{H_2SO_4}=0,016\left(mol\right)\)
\(\Rightarrow m_{Zn\left(pứ\right)}=n\cdot M=0,016\cdot32=0,512\left(g\right)\)
\(\Rightarrow m_{Zn\left(dư\right)}=m_{Zn\left(ĐB\right)}-n_{Zn\left(Pứ\right)}=1,95-0,512=1,438\left(g\right)\)
c, Theo PT : \(n_{H_2}=n_{H_2SO_4}=0,016\left(mol\right)\)
\(\Rightarrow V_{H_{2\left(đktc\right)}}=n\cdot22,4=0,016\cdot22,4=0,3584\left(l\right)\)
\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\
m_{H_2SO_4}=\dfrac{22,05.20}{100}=4,41\left(g\right)\\
n_{H_2SO_4}=\dfrac{4,41}{98}=0,045\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
\(LTL:\dfrac{0,03}{1}< \dfrac{0,045}{1}\)
=> H2SO4 dư
\(n_{H_2SO_4\left(p\text{ư}\right)}=n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,03\left(mol\right)\\
m_{H_2SO_4\left(d\right)}=\left(0,045-0,03\right).98=1,47\left(g\right)\\
m_{\text{dd}}=1,95+22,05-\left(0,03.2\right)=23,94\left(g\right)\\
C\%_{ZnCl_2}=\dfrac{0,03.136}{23,94}.100\%=17\%\)
\(a,n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ n_{H_2SO_4}=\dfrac{22,05}{98}=0,225\left(mol\right)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
bđ 0,03 0,225
pư 0,03 0,03
spư 0 0,195 0,03 0,03
\(b,m_{H_2SO_4\left(dư\right)}=0,195.98=19,11\left(g\right)\\ c,m_{dd}=1,95+22,05-0,03.2=23,94\left(g\right)\\ C\%_{ZnSO_4}=\dfrac{0,03.161}{23,94}.100\%=20,18\%\)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=147.10\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: 0,1 0,1 0,1 0,1
Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) ⇒ Zn hết, H2SO4 dư
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) mdd sau pứ = 6,5 + 147 - 0,1.2 = 153,3 (g)
\(C\%_{ddZnSO_4}=\dfrac{0,1.161.100\%}{153,3}=10,502\%\)
\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,15-0,1\right).98.100\%}{153,3}=3,196\%\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{18,56}{232}=0,08\left(mol\right)\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,08}{1}>\dfrac{0,2}{4}\), ta được Fe3O4 dư.
Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
\(n_{Zn}=\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\\ PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)
tỉ lệ 1 : 1 : 1 : 1
n(mol) 0,25-->0,25------->0,25------>0,25
\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\\ m_{ZnSO_4}=n\cdot M=0,25\cdot\left(65+32+16\cdot4\right)=40,25\left(g\right)\)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\), ta được Zn dư.
Theo PT: \(n_{Zn\left(pư\right)}=n_{H_2SO_4}=0,015\left(mol\right)\Rightarrow n_{Zn\left(dư\right)}=0,03-0,015=0,015\left(mol\right)\)
\(\Rightarrow m_{Zn\left(dư\right)}=0,015.65=0,975\left(g\right)\)
c, \(n_{H_2}=n_{H_2SO_4}=0,015\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,015.22,4=0,336\left(l\right)\)
\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)
a)\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,05 0,1 0,05 0,05
b)\(m_{ZnCl_2}=0,05\cdot136=6,8g\)
c)\(V_{H_2}=0,05\cdot22,4=1,12l\)
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\)
Phản ứng thế
\(b,n_{Zn}=\dfrac{1,3}{65}=0,02(mol)\\ \Rightarrow n_{ZnCl_2}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{ZnCl_2}=0,02.136=2,72(g)\\ V_{H_2}=0,02.22,4=0,448(l)\)
4.1)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + H2SO4 (loãng) ---> ZnSO4 + H2
0,1---->0,1---------------------------->0,1
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) \(m_{ddH_2SO_4}=\dfrac{0,1.98}{19,6\%}=50\left(g\right)\)
4.2)
Gọi kim loại hóa trị II là R; \(n_{AgCl}=\dfrac{2,87}{143,5}=0,02\left(mol\right)\)
PTHH: \(RCl_2+2AgNO_3\rightarrow R\left(NO_3\right)_2+2AgCl\downarrow\)
0,01<-----------------------------------0,02
=> \(M_{RCl_2}=\dfrac{0,95}{0,01}=95\left(g/mol\right)\)
=> \(M_R=95-71=24\left(g/mol\right)\)
Mà R có hóa trị II => R là Magie (Mg)
Câu 5:
a) Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Al_2O_3}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)
=> 80a + 102b = 14,2 (1)
nHCl = 0,2.3,5 = 0,7 (mol)
PTHH:
CuO + 2HCl ---> CuCl2 + H2O
a------>2a
Al2O3 + 6HCl ---> 2AlCl3 + 3H2O
b------->6b
b) 2a + 2b = 0,7 (2)
Từ (1), (2) => a = 0,05; b = 0,1
=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{14,2}.100\%=28,17\%\\\%m_{Al_2O_3}=100\%-28,17\%=71,83\%\end{matrix}\right.\)
a) Zn + H2SO4 --> ZnSO4 + H2
b) \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\) => Zn dư, H2SO4 hết
PTHH: Zn + H2SO4 --> ZnSO4 + H2
____0,015<-0,015--->0,015->0,015
=> mZn(dư) = (0,03-0,015).65 = 0,975 (g)
c) VH2 = 0,015.22,4 = 0,336(l)
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