Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
TTĐ:
\(m_{Fe_3O_4}=46,4\left(g\right)\)
\(C_{M_{H_2SO_4}}=2\left(M\right)\)
___________
a) \(V_{H_2SO_4}=?\left(l\right)\)
b)\(C_{M_{Fe_2\left(SO_4\right)_3}}=?\left(M\right)\)
Giải
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{46,4}{232}=0,2\left(mol\right)\)
\(Fe_3O_4+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+FeSO_4+4H_2O\)
\(0,2\rightarrow0,8\) : 0,2 : 0,2 (mol)
\(a)V_{H_2SO_4}=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4\left(l\right)\)
\(b)C_{M_{FeSO_{\text{4 }}}}=C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{n}{V}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
\(a)n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2mol\\Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)
0,2 0,8 0,2 0,2 0,8
\(V_{H_2SO_4}=\dfrac{0,8}{2}=0,4l\\ b)C_{M\left(FeSO_4\right)}=\dfrac{0,2}{0,4}=0,4M\\ C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,4}=0,5M\)
a, Ta có: \(n_{KOH}=0,15.2=0,3\left(mol\right)\)
PT: \(KOH+HCl\rightarrow KCl+H_2O\)
Theo PT: \(n_{HCl\left(pư\right)}=n_{KOH}=0,3\left(mol\right)\)
Mà: HCl dư 15%
\(\Rightarrow n_{HCl}=0,3+0,3.15\%=0,345\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,345}{1,5}=0,23\left(l\right)\)
b, Theo PT: \(n_{KCl}=n_{KOH}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{KCl}}=\dfrac{0,3}{0,15+0,23}\approx0,789\left(M\right)\)
\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,3.15\%}{0,15+0,23}\approx0,118\left(M\right)\)
1.nCO2=0,1 (mol )
TH1: Số mol của CO2 dư => Khối lượng muối khan tối đa tạo được là:
mmuối=0,1.84=8,4<9,5 (loại )
TH2: CO2 hết
Gọi số mol CO2 tạo muối Na2CO3;NaHCO3 lần lượt là x, y
2NaOH+CO2→Na2CO3+H2O
NaOH+CO2→NaHCO3
Ta có : \(\left\{{}\begin{matrix}x+y=0,1\\106x+84y=9,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
⇒nNaOH=2.0,05+0,05=0,15 (mol)
⇒CMNaOH=\(\dfrac{0,15}{0,1}\)=1,5M
300ml = 0,3l
\(n_{HNO3}=1.0,3=0,3\left(mol\right)\)
Pt : \(NaOH+HNO_3\rightarrow NaNO_3+H_2O|\)
1 1 1 1
0,3 0,3 0,3
\(n_{NaOH}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
200ml = 0,2l
\(C_{M_{NaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
\(n_{NaNO3}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{NaNO3}=0,3.85=25,5\left(g\right)\)
Sau phản ứng :
\(V_{dd}=0,2+0,3=0,5\left(l\right)\)
\(C_{M_{NaNO3}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
Chúc bạn học tốt
\(n_{HNO_3}=0,3\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
Theo PT: \(n_{NaOH}=n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)
\(m_{NaNO_3}=0,3.85=25,5\left(g\right)\)
a) \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
b) \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{CO_2}=0,1\left(mol\right)\)
=> \(CM_{NaOH}=\dfrac{0,1}{0,1}=1M\)
c) Sửa đề DNaOH = 1,2g/ml
\(m_{ddsaupu}=0,05.44+100.1,2=122,2\left(g\right)\)
\(n_{Na_2CO_3}=n_{CO_2}=0,05\left(mol\right)\)
=> \(C\%_{Na_2CO_3}=\dfrac{0,05.106}{122,2}.100=4,34\%\)
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)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)
Bài 1
\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)
Bài 5
\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
\(a)n_{H_2SO_4}=0,15.1=0,15mol\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,3 0,15 0,15 0,15
\(C_{M\left(NaOH\right)}=\dfrac{0,3}{0,1}=3M\\ b)C_{M\left(Na_2SO_4\right)}=\dfrac{0,15}{0,1+0,15}=0,6M\)