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!!!

Fe+2HCl→FeCl2+H2

Fe2O3+6HCl→2FeCl3+3H2O

nH2=2,24/22,4=0,1mol

→nFe=0,1mol

→mFe=0,1.56=5,6g

%mFe=5,6/10.100%=56%

%mFe2O3=100-56=44%

mFe2O3=10-5,6=4,4g

nFe2O3=4,4/160=0,0275mol

nHCl=0,1.2+0,0275.6=0,365mol

mHCl=0,365.36,5=13,3225g

C%HCl=13,3225/300.100=4,44%

mdd spứ=10+300-0,1.2=309,8g

mFeCl2=0,1.127=12,7g

C%FeCl2=12,7/309,8.100%=4,1%

mFeCl3=0,0275.2.162,5=8,9375g

C%FeCl3=8,9375/309,8.100%=2,88%