a) PTHH: AgNO3 + HCl ----> AgCl + HNO3
m\(AgNO_3\) = \(\dfrac{10\%.510}{100\%}=51\left(g\right)\)
=> n\(AgNO_3\) = \(\dfrac{51}{170}=0,3\left(mol\right)\)
mHCl = \(\dfrac{91,25.20\%}{100\%}=18,25\left(g\right)\)
b) nHCl = \(\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
Ta có tỉ lệ: \(\dfrac{0,3}{1}< \dfrac{0,5}{1}\) => AgNO3 phản ứng hết, HCl dư
Theo PTHH: nHCl(p/ứ) = n\(AgNO_3\) = 0,3 (mol)
=> nHCl(dư) = 0,5-0,3 = 0,2 (mol)
=> mHCl(dư) 0,2.36,5 = 7,3 (g)
c) Theo PTHH: n\(AgCl\) = n\(AgNO_3\) = 0,3 (mol)
=> m\(AgCl\) = 0,3.143,5 = 43,05 (g)
d) mddsp/ứ = 510 + 91,25 - 43,05 = 558,2 (g)
C%HCl(dư) = \(\dfrac{7,3}{558,2}.100\%=1,3\%\)
C%\(HNO_3\) = \(\dfrac{18,9}{558,2}.100\%=3,39\%\)
