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1.
a.\(n_{HCl}=0,2.0,15=0,03\left(mol\right)\)
b.\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
c.\(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\)
d.\(m_{H_2SO_4}=10\%.9,8=0,98\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{0,98}{98}=0,01\left(mol\right)\)
e.\(m_{NaOH}=6.5\%=0,3\left(g\right)\Rightarrow n_{NaOH}=\dfrac{0,3}{40}=0,0075\left(mol\right)\)
f.\(m_{ddNaOH}=125.1,2=150\left(g\right)\Rightarrow m_{NaOH}=150.20\%=30\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{30}{40}=0,75\left(mol\right)\)
2.
\(m_{NaOH}=10.20\%=2\left(g\right)\Rightarrow n_{NaOH}=\dfrac{2}{40}=0,05\left(mol\right)\)
PTHH: 2NaOH + H2SO4 → Na2SO4 + H2O
Mol: 0,05 0,025
\(\Rightarrow m_{Na_2SO_4}=0,025.142=3,55\left(g\right)\)
3.
\(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)
PTHH: 2NaCl + 2H2O → 2NaOH + Cl2 + H2
Mol: 0,1 0,1
\(m_{NaOH}=0,1.40=4\left(g\right)\Rightarrow m_{ddNaOH}=\dfrac{4.100\%}{5\%}=80\left(g\right)\)
\(\Rightarrow V_{ddNaOH}=\dfrac{80}{1,2}=66,7\left(ml\right)\)
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Zn + 2HCl => ZnCl2 + H2
Na2CO3 + 2HCl=> 2NaCl + H2O + CO2
MY = 0,5875.32 = 18,8
áp dụng sơ đồ đường chéo ta đc nH2 : nCO2 = 3:2
mà nH2 = nZn ; nCO2 = nNa2CO3
=> nZn = 3/2 nCO2
ta có \(65.\frac{3}{2}x+106x=4,07\left(g\right)\) => x= 0,02 mol => nZn =0,03
a. => % na2CO3 = \(\frac{0,02.106}{4,07}.100\%=52,088\%\)
=> % Zn = 47,912%
b. nHCl pư = 2 .nZn + 2. nNa2CO3 = 2.0,03+ 2.0,02 = 0,1
=> mHCl pư = 0,1.36,5 = 3,65 (g)
=> m HCl dùng = 3,65.120% = 4,38 (g)
=> mdd HCl = \(\frac{4,38.100}{25}=17,52\)
=> mdd = 4,07 + 17,52 - 0,03.2-0,02.44 = 20,65(g)
mHCl dư = 4,38 - 3,65 = 0,73(g)
C% HCl dư = \(\frac{0,73}{20,65}.100\%\) = 3,535%
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4FeS2+ 11O2 => 2Fe2O3 + 8SO2
2SO2 + O2 => 2SO3 (xúc tác v2O5)
SO3 + H2O => H2SO4
Fe2O3+ 3H2SO4 => Fe2(SO4)3 + 3H2O
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\(\Delta H=1.E_{C=C}+4.E_{C-H}+1.E_{H-H}-1.E_{C-C}-6.E_{C-H}\)
= 1.142,5 + 4.99 + 1.104,2 - 1.78 - 6.99 = -29,3 (kcal/mol)
=> A
dO2\H2=32\2=16
dN2\H2=28\2=14
dCO2\H2=44\2=22
B)
dO2\kk=32\29=1,1
dN2\kk=28\29=0,9655
dCO2\kk=44\29=1,517