Đáp án:
Giải thích các bước giải:
a)$n_{H_2}=0,1(mol)$
$Fe+2HCl→FeCl_2+H_2↑$
$0,1:0,2:0,1:0,1$
$m_{Fe}=0,1.56=5,6(g)$
%$m_{Fe}=\frac{5,6}{10}.100=56$%
%$m_{Fe_2O_3}=100$%$-56$%$=44$%
b)$n_{Fe_2O_3}=\frac{10-5,6}{160}=0,0275(mol)$
$2Fe_2O_3+6HCl→2FeCl_3+3H_2↑$
$0,0275:0,0,0825:0,0275:0,04125$
$∑m_{HCl}=36,5(0,2+0,0825)=10,31125(g)$
$C_{HCl}=\frac{10,31125}{300}.100≈3,437$%
c)$m_{ddspứ}=10+300-0,1.2=309,8(g)$
d)$C_{FeCl_2}=\frac{0,1.127}{309,8}.100≈4,1$%
$C_{FeCl_3}=\frac{0,0275.162,5}{309,8}.100≈1,442$%
Xin hay nhất!!!