mdd HCl=11,2+188,8=200(g)
\(\text{C%HCl=11,2/200x100=5,6%}\)
nHCl trong 97,768 g dd A=97,768x5,6%/36,5=0,15(mol)
\(\text{HCl+AgNO3->AgCl+HNO3}\)
\(\text{nAgNO3=114x15%/170=0,1(mol)}\)
=>HCl dư
mdd spu=114+97,768-0,1x143,5=197,418(g)
\(\text{C%hCl dư=0,05x36,5/197,418=0,924%}\)
\(\text{C% HNO3=0,1x63/197,418=3,19%}\)