дано
m(NH4CL) = 20 g
m(Ca(OH)2) = 20 g
----------------------
V(NH3)-?
M(NH4CL) = 53.5 g/mol
n(NH4CL) = m/M = 20 / 53.5 = 0.37 mol
M(Ca(OH)2) = 74g/mol
n(Ca(OH)2) = m/M = 20/74 = 0.27 mol
n(NH4CL) > n(Ca(OH)2)
2NH4CL+Ca(OH)2-->CaCL2+2NH3+2H2O
n(Ca(OH)2) = 2n(NH3)
n(NH3) = 2*0.27 = 0.54 mol
V(NH3) = n*Vm = 0.54 * 22.4 = 12.096L
ответ 12.096 л
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Answers & Comments
дано
m(NH4CL) = 20 g
m(Ca(OH)2) = 20 g
----------------------
V(NH3)-?
M(NH4CL) = 53.5 g/mol
n(NH4CL) = m/M = 20 / 53.5 = 0.37 mol
M(Ca(OH)2) = 74g/mol
n(Ca(OH)2) = m/M = 20/74 = 0.27 mol
n(NH4CL) > n(Ca(OH)2)
2NH4CL+Ca(OH)2-->CaCL2+2NH3+2H2O
n(Ca(OH)2) = 2n(NH3)
n(NH3) = 2*0.27 = 0.54 mol
V(NH3) = n*Vm = 0.54 * 22.4 = 12.096L
ответ 12.096 л