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a, \(HCl+NaOH\rightarrow NaCl+H_2O\)
b, \(n_{HCl}=0,3.2=0,6\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{24}{20\%}=120\left(g\right)\)
c, \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\\ n_{NaOH}=0,2.0,1=0,02\left(mol\right)\\ n_{CH_3COOH}=n_{NaOH}=0,02\left(mol\right)\\ C_{MddCH_3COOH}=\dfrac{0,02}{0,1}=0,2\left(M\right)\)
\(n_{NaOH}=0,2.1,5=0,3\left(mol\right)\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,1---------->0,1
=> mNaOH = 0,1.40 = 4 (g)
=> \(C\%_{NaOH}=\dfrac{4}{80}.100\%=5\%\)
PTHH: CH3COOH + C2H5OH --H2SO4(đặc), to--> CH3COOC2H5 + H2O
0,3------------------------------------------------>0,3
=> meste = 0,3.88.80% = 21,12 (g)
PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2CO_3}=0,3\left(mol\right)\\n_{NaOH}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{NaOH}=\dfrac{0,6\cdot40}{240}\cdot100\%=10\%\\C\%_{Na_2CO_3}=\dfrac{0,3\cdot106}{240+0,3\cdot44}\cdot100\%\approx12,56\%\end{matrix}\right.\)
\(n_{NaOH}=\dfrac{150.8\%}{40}=0,3\left(mol\right)\)
PTHH:
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,3 0,15 0,15 0,3
\(m_{HCl}=0,15.98=14,7\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,7.100}{4,9}=300\left(g\right)\)
\(b,m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)
\(c,C\%_{Na_2SO_4}=\dfrac{21,3}{150+300}.100\%=4,733\%\)
Phần tính m dd axit bị nhầm thành thành HCl rồi em nhé, dẫn tới phần c cũng sai theo.
a) 2Na+2H2O--->2NaOH+H2
n\(_{Na}=\frac{13,8}{23}=0,6\left(mol\right)\)
Theo pthh
n\(_{H2}=\frac{1}{2}n_{Na}=0,3\left(mol\right)\)
V\(_{H2}=0,3.22,4=6,72\left(l\right)\)
b) Theo pthh
n\(_{NaOH}=n_{Na}=0,6\left(mol\right)\)
m\(_{NaOH}=0,6.40=24\left(g\right)\)
c) C\(_{M\left(NaOH\right)}=\frac{0,6}{0,1}=6\left(M\right)\)
\(\text{Na + H2O -> NaOH + 1/2H2}\)
\(\text{a) Ta có: n Na=13,8/23=0,6 mol}\)
Theo ptpu: nH2=1/2 nNa=0,3 mol
\(\Rightarrow\text{ V H2=0,3.22,4=6,72 lít}\)
b) Theo ptpu: nNaOH=nNa=0,6 mol
\(\Rightarrow\text{mNaOH=0,6.40=24 gam}\)
\(\text{c) Ta có V dung dịch sau phản ứng=100 ml =0,1 lít}\)
\(\Rightarrow\text{CM NaOH =nNaOH/V dung dịch=0,6/0,1=6M}\)
a, \(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Theo PT: \(n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\)
\(\Rightarrow a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\)
b, \(m_{dd}=\dfrac{15}{2\%}=750\left(g\right)\)
\(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\\ NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\\a, n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\\ a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\\ b,m_{ddCH_3COOH}=\dfrac{15.100}{2}=750\left(g\right)\)
\(n_{CuSO_4}=\dfrac{20}{160}=0,125(mol)\\ a,CuSO_4+2NaOH\to Cu(OH)_2\downarrow+2NaCl\\ \Rightarrow n_{NaOH}=0,25(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{0,25}{0,2}=1,25M\\ b,n_{Cu(OH)_2}=0,125(mol)\\ \Rightarrow m_{Cu(OH)_2}=0,125.98=12,25(g)\\ c,Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=0,125(mol)\\ \Rightarrow m_{CuO}=0,125.80=10(g)\)
H2SO4 + 2NaOH \(\rightarrow\) Na2SO4 + H2O
a) Ta có: m H2SO4=120.10%=12 gam \(\rightarrow\) nH2SO4=\(\frac{12}{98}\) mol
Theo ptpu: nNa2SO4=nNa2SO4=12/98 mol
\(\rightarrow\)mNa2SO4=\(\frac{12}{98}\) .(23.2+96)=17,39 gam
b)Ta có: nNaOH=2nH2SO4=\(\frac{12}{49}\) mol
\(\rightarrow\) V NaOH=\(\frac{12}{49}\) :2=\(\frac{6}{49}\) lít