Đáp án:
Giải thích các bước giải:
a, $n_{H2}$ =$\frac{2,24}{22,4}$ =0,1 mol
PT:Mg+2HCl→MgCl2+H2
0,1→ 0,2 → 0,1 → 0,1 mol
$m_{Mg}$ =0,1.24=2,4 (g)
% $m_{Mg}$ =$\frac{2,4}{10}$ .100=24%
% $m_{Cu}$ =100-24=76 %
b, $C_{MHCl}$ =$\frac{0,2}{0,3}$ =0,67 M
Mg+2HCl→MgCl2+H2
nH2=2,24/22,4=0,1mol
mMg=0,1.24=2,4g
%mMg=2,4/10.100%=24%
%mCu=100-24=76%
nHCl=0,1.2=0,2mol
CM HCl=0,2/0,3=0,67M
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