$PTHH:Fe+CuSO_4\to FeSO_4+Cu\downarrow$

Đặt $n_{Cu}=x(mol)\Rightarrow n_{Fe}=x(mol)$

$m_{KL tăng}=m_{Cu}-m_{Fe}=26,4-20$

$\Rightarrow 64x-56x=6,4$

$\Rightarrow x=0,8$

$\Rightarrow m_{Cu}=0,8.64=51,2(g)$

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$

$\Rightarrow n_{Al}=0,15(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$

$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$

$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$

$3Fe_2+2O_2\xrightarrow{t^o}Fe_3O_4$

Theo PT: $n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}.0,15=0,05(mol)$

vô tcn của ng e mún tặng xong r ở bên trái cái avt có ô tặng coin á

$a)Ba(NO_3)_2+H_2SO_4\to BaSO_4\downarrow+2HNO_3$

$CaCO_3+H_2SO_4\to CaSO_4\downarrow+H_2O+CO_2\uparrow$

$BaSO_3+H_2SO_4\to BaSO_4\downarrow+H_2O+SO_2\uparrow$

$b)Mg+H_2SO_4\to MgSO_4+H_2\uparrow$

$c)CaCO_3+H_2SO_4\to CaSO_4\downarrow+H_2O+CO_2\uparrow$

$d)BaSO_3+H_2SO_4\to BaSO_4\downarrow+H_2O+SO_2\uparrow$

$e)2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O$

Kí hiệu khí, kết tủa có trong PT

Câu e: $Fe_2(SO_4)_3$ là dd đỏ nâu

là sao em :)

Bài 5:

$a)2KClO_3\xrightarrow{t^o}2KCl+3O_2$

$b)n_{KClO_3}=\dfrac{490}{122,5}=4(mol)$

Theo PT: $n_{O_2}=1,5n_{KClO_3}=6(mol)$

$\Rightarrow V_{O_2(đktc)}=6.22,4=134,4(lít)$

Bài 6:

$a)2Al(OH)_3+3H_2SO_4\to Al_2(SO_4)_3+6H_2O$

$n_{Al(OH)_3}=\dfrac{78}{78}=1(mol)$

Theo PY: $n_{H_2SO_4}=1,5(mol);n_{Al_2(SO_4)_3}=0,5(mol)$

$\Rightarrow m_{H_2SO_4}=1,5.98=147(g)$

$m_{Al_2(SO_4)_3}=0,5.342=171(g)$

Bài 3:

$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$

$Fe+2HCl\to FeCl_2+H_2\uparrow$

Theo PT: $n_{FeCl_2}=n_{H_2}=0,2(mol)$

$\Rightarrow m_{FeCl_2}=0,2.127=25,4(g);V_{H_2}=0,2.22,4=4,48(lít)$

$\Rightarrow m=25,4;V=4,48$

Bài 4:

$CuO+H_2SO_4\to CuSO_4+H_2O$

Theo PT; $n_{H_2SO_4}=n_{CuSO_4}=n_{CuO}=\dfrac{32}{80}=0,4(mol)$

$\Rightarrow m_{H_2SO_4}=0,4.98=39,2(g)$

$m_{CuSO_4}=0,4.160=64(g)$

$PTHH:NaOH+HCl\to NaCl+H_2O$

$n_{NaOH}=\dfrac{4}{40}=0,1(mol)$

Tỉ lệ: $n_{NaOH}<n_{HCl}\Rightarrow HCl$ dư

Theo PT: $n_{NaCl}=n_{NaOH}=0,1(mol)$

$\Rightarrow m_{muối}=m_{NaCl}=0,1.58,5=5,85(g)$