Дано
m(ppaBaCL2) =208 g
W(BaCL2) = 15%
-------------------------
V(HCL)-?
m(BaCL2) = 208*15% / 100% = 31.2 g
Ba(OH)2+2HCL-->BaCL2+2H2O
M(BaCL2) = 208 g/mol
n(BaCL2) = m/M = 31.2 / 208 = 0.15 mol
n(HCL) = 2*0.15 = 0.3 mol
V(HCL) = n*Vm = 0.3 * 22.4 = 6.72 L
ОТВЕТ 6.72 л
1)=CuCl2+BaSO4 2)=Li3PO4+H2O
Br2 + 5Cl2 + 6H2O ---t--> 2HBrO3 + 10HCl
HBrO3 + XeF2 + H2O ---> HBrO4 + 2HF + Xe