дано
m(HCL) = 14,6 g
η(CO2) = 90%
Vпр (CO2)-?
Na2CO3+2HCL-->2NaCL+H2O+CO2
M(HCL) = 36.5 g/moL
n(HCL) = m/M = 14.6 / 36.5 = 0.4 mol
2n(HCL) = n(CO2)
n(CO2) = 0.4 / 2 = 0.2 mol
V теор(CO2) = n(CO2) * Vm = 0.2*22.4 = 4.48 L
V пр (CO2) = 4.48 * 90% / 100% = 4.032 L
ответ 4.032 л
Объяснение:
дано
m(HCL) = 14,6 g
η(CO2) = 90%
Vпр (CO2)-?
Na2CO3+2HCL-->2NaCL+H2O+CO2
M(HCL) = 36.5 g/moL
n(HCL) = m/M = 14.6 / 36.5 = 0.4 mol
2n(HCL) = n(CO2)
n(CO2) = 0.4 / 2 = 0.2 mol
V теор(CO2) = n(CO2) * Vm = 0.2*22.4 = 4.48 L
V пр (CO2) = 4.48 * 90% / 100% = 4.032 L
ответ 4.032 л
дано
m(HCL) = 14,6 g
η(CO2) = 90%
Vпр (CO2)-?
Na2CO3+2HCL-->2NaCL+H2O+CO2
M(HCL) = 36.5 g/moL
n(HCL) = m/M = 14.6 / 36.5 = 0.4 mol
2n(HCL) = n(CO2)
n(CO2) = 0.4 / 2 = 0.2 mol
V теор(CO2) = n(CO2) * Vm = 0.2*22.4 = 4.48 L
V пр (CO2) = 4.48 * 90% / 100% = 4.032 L
ответ 4.032 л
Объяснение: