Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
\(PTHH:2NaOH+H_2S\rightarrow Na_2S+2H_2O\)
_________________0,05______0,05____________
\(NaOH+H_2S\rightarrow NaHS+H_2O\)
_________0,1 ______0,1________
\(\Rightarrow\Sigma n_{H2S}=0,1+0,05=0,15\left(mol\right)\)
\(\Rightarrow V_{H2S}=0,15.22,4=3,36\left(l\right)\)
Câu 2:
Ta có:
\(n_{NaOH}=0,35.0,1=0,035\left(mol\right)\)
\(n_{H2S}=\frac{0,448}{22,4}=0,02\left(mol\right)\)
\(\Rightarrow T=\frac{n_{NaOH}}{n_{H2S}}=\frac{0,035}{0,02}=1,75\)
\(1< T< 2\Rightarrow\) Tạo cả 2 muối
\(2NaOH+H_2S\rightarrow Na_2S+2H_2O\)
0,035_____0,0175___0,0175_____ \(Na_2S+H_2S\rightarrow2NaHS\) 0,0025___0,0025___0,005 \(\Rightarrow\left\{{}\begin{matrix}CM_{Na2S}=\frac{0,0175-0,0025}{0,1}=0,15M\\CM_{NaHS}=\frac{0,005}{0,1}=0,05M\end{matrix}\right.\)Câu 3:
Ta có:
\(\left\{{}\begin{matrix}n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\\n_S=\frac{4}{32}=0,125\left(mol\right)\end{matrix}\right.\)
\(PTHH:Zn+S\rightarrow ZnS\)
Tỉ lệ : \(\frac{0,2}{1}>\frac{0,125}{1}\Rightarrow\) Zn dư
\(n_{ZnS}=0,125\left(mol\right)\)
\(n_{Zn\left(Dư\right)}=0,2-0,125=0,075\left(mol\right)\)
\(ZnS+2HCl\rightarrow ZnCl_2+H_2S\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\left\{{}\begin{matrix}n_{H2S}=n_{ZnS}=0,125\left(mol\right)\\n_{H2}=n_{Zn}=0,075\left(mol\right)\end{matrix}\right.\)
\(M_{hh}=\frac{0,125.34+0,075.2}{0,2}=22\)
\(D_{hh/H2}=\frac{22}{2}=11\)
a, PT: \(Fe+S\underrightarrow{t^o}FeS\)
Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,1}{1}\), ta được Fe dư.
Chất rắn A gồm Fe dư và FeS.
Theo PT: \(n_{Fe\left(pư\right)}=n_{FeS}=n_S=0,1\left(mol\right)\)
\(\Rightarrow n_{Fe\left(dư\right)}=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Fe\left(dư\right)}=0,2\left(mol\right)\\n_{H_2S}=n_{FeS}=0,1\left(mol\right)\end{matrix}\right.\)
Ở cùng điều kiện nhiệt độ và áp suất, %V cũng là % số mol.
\(\Rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,2}{0,2+0,1}.100\%\approx66,67\%\\\%V_{H_2S}\approx33,33\%\end{matrix}\right.\)
b, Ta có: \(\Sigma n_{HCl\left(dadung\right)}=2n_{Fe}+2n_{FeS}=0,6\left(mol\right)\) (1)
PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(n_{NaOH}=0,1.2=0,2\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=n_{NaOH}=0,2\left(mol\right)\) (2)
Từ (1) và (2) \(\Rightarrow\Sigma n_{HCl}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,8}{0,5}=1,6\left(M\right)\)
Bạn tham khảo nhé!
B1:
2NaOH+H2SO4\(\rightarrow\)Na2SO4+2H2O
nNaOH=\(\frac{4}{40}=0.1\)mol
=>nH2SO4=\(\frac{1}{2}\)nNaOH=0.05 mol
=>CM=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10-4 (M)
B2:
Mg+\(\frac{1}{2}\)O2\(\underrightarrow{t^0}\)MgO (1)
MgO+2HCl\(\rightarrow\)MgCl2+H2O (2)
nMg(1)=\(\frac{0,36}{24}=0,015mol\)
=>nMgO(1)=0,015=nMgO(2)
nHCl(2)=2nMgO(2)=0,03mol
=>CM(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)
a,\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,05 0,1 0,05 0,05
PTHH: MgO + 2HCl → MgCl2 + H2O
Mol: 0,1 0,2 0,1
⇒ mMg = 0,05.24 = 1,2 (g)
mMgO = 5,2 - 1,2 = 4 (g)
b,\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
⇒ nHCl đã dùng = 0,1+0,2 = 0,3 (mol)
\(C_{M_{ddHCl}}=\dfrac{0,3}{0,5}=0,6\left(l\right)\)
c,\(C_{M_{MgCl_2}}=\dfrac{0,05+0,1}{0,6}=0,25M\)
a) Gọi x là số mol Na2CO3
y là số mol MgCO3
Ta có PT:
Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O
....x..............2x............2x.........x...........x
MgCO3 + 2HCl ---> MgCl2 + CO2 + H2O
....y............2y................y...........y.........y
Theo bài ra ta có:
106x + 84y = 3,96(1)
\(n_{CO_2}\)= \(\frac{0,896}{22,4}\)=0,04(mol)
<=> x+y= 0,4(2)
Giải 1 và 2 ta có:
x=0,015 ➝mNaCl = 0,015. 58,5 = 0,8775(g)
y=0,025 ➝m\(MgCl_2\)= 0,025.95=2,375(g)
mHCl= \(\frac{3,65.100}{100}\)=3,65(g)
nHCl = \(\frac{3,65}{36.5}\)= 0,1(mol)
=> nHCl dư = 0,1 - (2x+2y) = 0,02(mol)
mHCl dư = 0,02 . 36,5 = 0,73(g)
Áp dụng ĐLBTKL ta có:
mx + mHCl = mY + m\(CO_2\)
=> my = 3,69 + 100 - 0,04.44= 101,93(g)
C%\(NaCl\)=\(\frac{0,8775}{101,93}.100\%\)=0,86%
C%\(MgCl_2\)=\(\frac{2,375}{101,93}.100\%\)=2,33%
C%HCl dư= \(\frac{0,73}{101,93}.100\%\)=0,72%
b) Ta có PT
MgCl2 + 2NaOH ---> Mg(OH)2 + 2NaCl
...0,025.....0,05
HCl + NaOH ---> NaCl + H2O
..0,02......0,02
=> VNaOH 0,35M = \(\frac{0,05+0,02}{0,35}\)= 0,2 lít
\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)
\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)
