ta có: nZn= 26/ 65= 0,4( mol)
PTPU
Zn+ 2HCl\(\rightarrow\) ZnCl2+ H2 (1)
0,4...0,8..........................
ta có mHCl= 600. 7,3%= 43,8( g)
nHCl= 43,8/ 36,5= 1,2( mol)> 0,8 mol
\(\Rightarrow\) HCl dư 0,4 mol
theo(1)=>nH2= nZnCl2= nZn= 0,4( mol)
ta có: mdd sau pư= mZn+ mdd HCl- mH2
= 26+ 600- 0,4. 2
= 625,2( g)
\(\Rightarrow\) C%HCl= \(\dfrac{0,4.36,5}{625,2}\). 100%\(\approx\) 2,34%
C%ZnCl2= \(\dfrac{0,4.136}{625.2}\). 100%\(\approx\) 8,7%
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