Đáp án:
b) 28,1 g
Giải thích các bước giải:
a/
$Ba+2HCl→BaCl_2+H_20,1$
$Ba+2H_2O→Ba(OH)_2+H_2$
$CuSO_4+Ba(OH)_2→Cu(OH)_2↓+BaSO_4↓$
$Cu(OH)_2→CuO+H_2O$
b/
$\hspace{1,0cm}Ba+\hspace{0,2cm}2HCl→BaCl_2+H_2↑\\BĐ:0,1\hspace{1cm}0,12\\PƯ:0,06←0,12\\CL:\hspace{0cm}0,04\hspace{1cm}0$
$\hspace{1cm}Ba+\hspace{0,4cm}2H_2O→Ba(OH)_2+H_2\\\hspace{1cm}0,06→\hspace{2,5cm}0,06$
$\hspace{0,7cm}CuSO_4+Ba(OH)_2→Cu(OH)_2↓+BaSO_4↓\\BĐ:0,1\hspace{1,4cm}0,06\\PƯ:0,06\hspace{0,4cm}←0,06→\hspace{1cm}0,06\hspace{2cm}0,06\\CL:0,04\hspace{1,5cm}0\hspace{2,1cm}0,06\hspace{2cm}0,06$
Nung nóng chất rắn chỉ xảy ra đối với: $Cu(OH)_2$
$Cu(OH)_2\xrightarrow{t^o}CuO+H_2O$
$⇒m=m_{BaSO_4}+m_{CuO}=0,1.233+0,06.80=28,1\ g$
