Đáp án:
%$m_{Al}$ = $49,1_{}$%
%$m_{Cu}$ = $50,9_{}$%
Giải thích các bước giải:
a) PTHH: $Cu+H_{2}SO_4$ --x-->
PTHH: $2Al+3H_{2}SO_4$ → $Al_{2}(SO_4)_3$ + $3H_{2}$ ↑
Hspt: 2 3 1 3 (mol)
Pư: 0,1 0,15 0,05 0,15 (mol)
Tóm tắt:
$m_{hh}$ = $m_{Cu}$ + $m_{Al}$ = $5,5(g)_{}$
$V_{H_2}=3,36(l)$
%$m_{Al}=?$ $(_{}$%$)_{}$
%$m_{Cu}=?$ $(_{}$%$)_{}$
Tính số mol: $n_{H_2}$ = $\frac{V_{H_{2} } }{22,4}$ = $\frac{3,36}{22,4}$ = $0,15(mol)_{}$
Tính theo yêu cầu:
b) $m_{Al}$ = $n_{Al}$ . $M_{Al}$
= $0,1.27=2,7(g)_{}$
%$m_{Al}$ = $m_{Al}.100$% $/ m_{hh}$
= $2,7.100_{}$% $/5,5_{}$
≈ $49,1_{}$%
%$m_{Cu}$ = $100_{}$% - $m_{Al}$ = $100_{}$% - $49,1_{}$% = $50,9_{}$%