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\(CM_{HCl\left(sau\right)}=\dfrac{n}{V}=\dfrac{0,1.2+0,2.2}{0,1+0,2}=2M\)
Ta có: \(n_{Cu\left(NO_3\right)_2}=0,2.1,5=0,3\left(mol\right)\)
PT: \(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_{2\downarrow}+2NaNO_3\)
_______0,3_______0,6_______0,3_________0,6 (mol)
a, mCu(OH)2 = 0,3.98 = 29,4 (g)
b, \(V_{ddNaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)
c, \(C_{M_{NaNO_3}}=\dfrac{0,6}{0,2+0,3}=1,2M\)
Bạn tham khảo nhé!
a) \(n_{Cu\left(NO_3\right)_2}=1,5.0,2=0,3\left(mol\right)\)
\(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
\(n_{Cu\left(OH\right)_2}=n_{Cu\left(NO_3\right)_2}=0,3\left(mol\right)\)
=> \(m_{Cu\left(OH\right)_2}=29,4\left(g\right)\)
b) \(n_{NaOH}=2n_{Cu\left(OH\right)_2}=0,6\left(mol\right)\)
=> \(V_{NaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)
c) \(CM_{NaCl}=\dfrac{0,3.2}{0,2+0,3}=1,2M\)
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)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)
\(NaOH+HCl-->NaCl+H2O\)
\(n_{NaOH}=0,2.2=0,4\left(mol\right)\)
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
==>NaOH dư.
dd sau pư là NaOH dư và NaCl
\(V_{dd}=200+100=300ml=0,3l\)
\(n_{NaOH}=n_{HCl}=0,2\left(mol\right)\)
\(n_{NaOH}dư=0,4-0,2=0,2\left(mol\right)\)
\(C_{M\left(NaOH\right)dư}=\frac{0,2}{0,3}=\frac{2}{3}\left(M\right)\)
\(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
\(C_{M\left(NaCl\right)}=\frac{0,2}{0,3}=\frac{2}{3}\left(M\right)\)