Ta có: \(n_{CaCl_2}=\dfrac{2,22}{111}=0,02\left(mol\right)\)
PT: \(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl_{\downarrow}\)
____0,02________________0,02_______0,04 (mol)
a, m chất rắn = mAgCl = 0,04.143,5 = 5,74 (g)
b, m muối = mCa(NO3)2 + mAgCl = 0,02.164 + 5,74 = 9,02 (g)
Bạn tham khảo nhé!