Theo mình nghĩ thu FeCl₂

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)

\(n_{HCl}=\frac{10,95}{36,5}=0,3\left(mol\right)\)

Theo PTHH: \(\frac{0,1}{1}< \frac{0,3}{2}\)=> Fe hết

\(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)

\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

\(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(a=12,7\left(g\right);b=2,24\left(l\right)\)

4) \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(n_{Al}=\frac{32,4}{27}=1,2\left(mol\right)\)

\(n_{O_2}=\frac{21,504}{22,4}=0,96\left(mol\right)\)

Theo PTHH, ta có: \(\frac{1,2}{4}< \frac{0,96}{3}\)=> O₂ dư

\(n_{O_2\left(dư\right)}=0,96-\left(\frac{1,2.3}{4}\right)=0,06\left(mol\right)\)

\(m_{O_2\left(dư\right)}=0,06.32=1,92\left(g\right)\)

b) \(n_{Al_2O_3}=\frac{2}{4}.n_{Al}=\frac{2}{4}.1,2=0,6\left(mol\right)\)

\(m_{Al_2O_3}=0,6.102=61,2\left(g\right)\)

c) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{H_2}=\frac{3}{2}.n_{Al}=\frac{3}{2}.1,2=1,8\left(mol\right)\)

\(V_{H_2}=1,8.22,4=40,32\left(l\right)\)

a) PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b) \(n_P=\frac{6,2}{31}=0,2\left(mol\right)\)

\(n_{P_2O_5}=\frac{2}{4}.n_P=\frac{2}{4}.0,2=0,1\left(mol\right)\)

\(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

Đặt CTHH: A₂O₃

Vì A chiếm 70% khối lượng nên oxi chiếm 30% khối lượng

\(m_{A_2O_3}=\frac{48.100}{30}=160\left(g\right)\)

\(2A+48=160\Rightarrow A=56\left(Fe\right)\)

CTHH: Fe₂O₃

Chọn C

a) Ban đầu có màu đen, sau chuyển thành màu đỏ gạch

b) \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

\(n_{CuO}=\frac{20}{80}=0,25\left(mol\right)\)

Chỉ thu được 16,8g chất rắn nên CuO không phản ứng hết.

Đặt n_CuO phản ứng là x, số mol CuO dư là 0,25-x

\(\left(0,25-x\right)80+64x=16,8\)

\(\Rightarrow x=0,2\)

\(H=\frac{0,2}{0,25}.100\%=80\%\)

c) \(n_{H_2}=n_{CuO\left(pứ\right)}=0,2\left(mol\right)\)

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

1) \(2Mg+O_2\underrightarrow{t^o}2MgO\)

a) \(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)

\(n_{O_2}=\frac{8}{32}=0,25\left(mol\right)\)

Theo PTHH, ta có: \(\frac{0,1}{2}< \frac{0,25}{1}\)=> O₂ dư

\(n_{O_2\left(dư\right)}=0,25-\left(\frac{0,1.1}{2}\right)=0,2\left(mol\right)\)

\(m_{O_2\left(dư\right)}=0,2.32=6,4\left(g\right)\)

\(n_{MgO}=n_{Mg}=0,1\left(mol\right)\)

\(m_{MgO}=0,1.40=4\left(g\right)\)