дано
m(ppa HCL) = 60 g
w(HCL) = 15%
V(CO2)-?
m(HCL) = m(ppa HCL) * W(HCL) / 100% = 60*15% / 100% = 9 g
Na2CO3+2HCL-->2NaCL+H2O+CO2
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 9/36 = 0.25 mol
n(HCL) = n(CO2) = 0.25 mol
V(CO2) = n(CO2)*Vn = 0.25*22.4 = 5.6 L
ответ 5.6 л
дано
m(ppa HCL) = 60 g
w(HCL) = 15%
V(CO2)-?
m(HCL) = m(ppa HCL) * W(HCL) / 100% = 60*15% / 100% = 9 g
Na2CO3+2HCL-->2NaCL+H2O+CO2
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 9/36 = 0.25 mol
n(HCL) = n(CO2) = 0.25 mol
V(CO2) = n(CO2)*Vn = 0.25*22.4 = 5.6 L
ответ 5.6 л