W(K)=M(K)\M(K2MnO4)= 78\197= 39,59%
w(K)=M(K)\M(KOH)=39\56=69,64%
w(K)=M(K)\M(K2O)=78\94=82,98%
1) C2H5OH = C2H4 + H2O
2) C2H4 + H2 = C2H6
C2H6 + Cl2 = C2H5Cl
3) C2H5Cl + NaOH = C2H5OH + NaCl
1)CaC2 + 2H2O = Ca(OH)2 + C2H2
2)2C2H2 + 5O2 = 4CO2 + 2H2O
3)CO2 + CaO = CaCO3
Ответ:
дано
m(CH3Br) = 3.85 g
m(Na) = 1.15 g
V пр (C2H6) = 385.4 ml = 0.3854 L
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η(C2H6)-?
2CH3Br+2Na-->C2H6+2NaBr
M(CH3Br) = 95 g/mol
n(CH3Br) = m/M = 3.85 / 95 = 0.04 mol
M(Na)= 23 g/mol
n(Na)= m/M = 1.15 / 23 = 0.05 mol
n(CH3Br) < n(Na)
2n(CH3Br) = n(C2H6)
n(C2H6) = 0.04 / 2 = 0.02 mol
Vтеор(C2H6) = n(C2H6) * Vm = 0.02 * 22.4 = 0.448 L
η(C2H6) = V пр(C2H6) / V теор(C2H6) * 100% = 0.3584 / 0.448 * 100% = 80%
ответ 80%
Объяснение: