Question

Помогите, пожалуйста

Answer 1

дано
m(ppa NH4CL) = 267.5 g
W(NH4CL) = 10%
--------------------------------
V(NH3)-?

m(NH4CL) = 267.5*10% / 100% = 26.75 g

NH4CL+NaOH-->NaCL+NH3+H2O
M(NH4CL) = 53.5 g/mol
n(NH4CL) = m/M = 26.75 / 53.5 = 0.5 mol
n(NH4CL)= n(NH3) = 0.5 mol
V(NH3) = n*Vm = 0.5*22.4 = 11.2 L
ответ 11.2 л