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\(m_{CH_3COOH}=\dfrac{80.9}{100}=7,2\left(g\right)\)
\(n_{CH_3COOH}=\dfrac{7,2}{60}=0,12\left(mol\right)\)
PTHH :
\(15CH_3COOH+10NaHCO_3\rightarrow10CH_3COONa+2H_2O+20CO_2\uparrow\)
0,12 0,08 0,08 0,016 0,16
\(a,m_{NaHCO_3}=84.0,08=6,72\left(g\right)\)
\(m_{ddNaHCO_3}=\dfrac{6,72.100}{4,2}=160\left(g\right)\)
\(b,m_{CH_3COONa}=0,08.82=6,56\left(g\right)\)
\(m_{H_2O}=0,016.18=0,288\left(g\right)\)
\(m_{CO_2}=0,16.44=7,04\left(g\right)\)
\(m_{ddCH_3COONa}=80+160-0,288-7,04=232,672\left(g\right)\)
\(C\%=\dfrac{6,56}{232,672}\approx2,82\%\)
\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,4 0,4 0,4 0,4
a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)
b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)
\(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)
\(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)
\(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)
c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)
0,4 0,3 0,3 0,3
\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)
PT \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+H_2O+CO_2\uparrow\)
a, \(m_{CH_3COOH}=\frac{6\times150}{100}=9\left(g\right)\)\(\Rightarrow n_{CH_3COOH}=\frac{9}{60}=0,15\left(mol\right)\)
Theo PT \(n_{NaHCO_3}=n_{CH_3COOH}=0,15\left(mol\right)\Rightarrow m_{NaHCO_3}=0,15\times84=12,6\left(g\right)\)\(\Rightarrow m_{ddNaHCO_3}=\frac{12,6\times100}{8,4}=150\left(g\right)\)
b, Theo PT \(n_{CH_3COONa}=n_{CH_3COOH}=0,15\left(mol\right)\)
\(\Rightarrow m_{CH_3COONa}=0,15\times82=12,3\left(g\right)\)
Có \(m_{ddsaupu}=150+150=300\left(g\right)\)
\(\Rightarrow C\%=\frac{12,3}{300}\times100=4,1\%\)
a. PTHH: \(CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
b. Đổi 100ml = 0,1 lít
Ta có: \(n_{Cu\left(OH\right)_2}=\dfrac{9,8}{98}=0,1\left(mol\right)\)
Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)
=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)
c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)
=> \(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)
Ta có: \(n_{NaOH}=0,1.0,5=0,05\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{CH_3COONa}=n_{NaOH}=0,05\left(mol\right)\)
a, \(C_{M_{CH_3COOH}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
b, \(m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)
a) \(n_{HCl}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
1 2 1 1 (mol)
0,2 0,4 0,2 0,2 (mol)
b) Thể tích khí hidro:
V = n.22,4 = 0,2.22,4 = 4,48 (l)
c) Khối lượng muối tạo thành:
\(m_{ZnCl_2}=n.M=0,2.\left(65+35,5.2\right)=27,2\left(g\right)\)
d) \(m_{ctHCl}=n.M=0,4.\left(1+35,5\right)=14,6\left(g\right)\)
\(C\%_{HCl}=\dfrac{m_{ctHCl}}{m_{ddHCl}}.100\%=\dfrac{14,6}{200}.100\%=7,3\%\)
a)
$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$
b)
n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)
m dd NaHCO3 = 0,2.84/8% = 210(gam)
c)
n CO2 = n CH3COOH = 0,2(mol)
=> V CO2 = 0,2.22,4 = 4,48(lít)
d)
m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100 + 210 - 0,2.44 = 301,2(gam)
C% CH3COONa = 0,2.82/301,2 .100% = 5,44%