Дано
m(ppa HCL) = 200 g
W(HCL) = 5%
m(K2CO3) = 17.25 g
----------------------------------
m(KCL)-?
m(HCL) = m(ppa HCL) * W(HCL) / 100% = 200*5 / 100 = 10 g
M(HCL) = 36. 5 g/mol
n(HCL) =m/M = 10/36.5 = 0.27 mol
M(K2CO3) = 138 g/mol
n(K2CO3) = m/M = 0.125 mol
n(HCL) > n(K2CO3)
M(KCL) = 74.5 g/mol
17.25 Xg
2HCL+K2CO3-->2KCL+H2O+CO2
138 2*74.5
X = 17.25 * 2*74.5 / 138 = 18.63 g
ответ 18.63 г