Bài 2:

$\rm a)\\n_{Fe}=\dfrac{5,6}{56}=0,1\ (mol)\\n_{HCl}=0,1×1=0,1\ (mol)$

Phương trình phản ứng:

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

Tỉ lệ: $\rm\dfrac{n_{Fe}}{1}>\dfrac{n_{HCl}}{2}\to Fe$ dư, tính theo HCl.

Theo phương trình:

$\rm n_{H_2}=n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,05\ (mol)$

$\rm V_{H_2}=0,05×24,79=1,2395\ (L)$

$\rm b)$

$\rm C_{M\ FeCl_2}=\dfrac{0,05}{0,1}=0,5\ (M)$

Bài 4:

$\rm a)$

Phương trình phản ứng:

$\rm 2Al+6HCl\to 2AlCl_3+3H_2$

$\rm b)\\n_{H_2}=\dfrac{7,437}{24,79}=0,3\ (mol)$

Theo phương trình:

$\rm n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\ (mol)\\n_{HCl}=2n_{H_2}=0,6\ (mol)$

$\rm\to a=0,2×27=5,4\ (g)\\\to b=\dfrac{0,6}{0,4}=1,5\ (M)$

Bài 5:

$\rm a)$

Phương trình phản ứng:

$\rm CuO+2HCl\to CuCl_2+H_2O$

$\rm b)\\n_{HCl}=\dfrac{50×3,65\%}{36,5}=0,05\ (mol)$

Theo phương trình:

$\rm n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{HCl}=0,025\ (mol)$

$\rm m_{CuO}=0,025×80=2\ (g)$

$\rm c)$

Bảo toàn khối lượng:

$\rm m_{dd\ CuCl_2}=2+50=52\ (g)$

$\rm C\%_{CuCl_2}=\dfrac{0,025×135}{52}×100\%=6,5\%$

Bài 2:

$a)\\ Fe+2HCl\to FeCl_2+H_2\\ n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ n_{HCl}=\dfrac{100}{1000}.1=0,1(mol)\\ n_{Fe}>\dfrac{n_{HCl}}{2}\to Fe\ du\ spu\\ n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05(mol)\\ V_{H_2}=0,05.24,79=1,2395(l)\\ b)\\ n_{FeCl_2}=n_{H_2}=0,05(mol)\\ C_{M\ dd\ FeCl_2}=\dfrac{0,05}{\frac{100}{1000}}=0,5(M)$

Bài 4:

$a)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ b)\\ n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)\\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)\\ a=m_{Al}=0,2.27=5,4(g)\\ n_{HCl}=2n_{H_2}=0,6(mol)\\ b=C_{M\ dd\ HCl}=\dfrac{0,6}{\frac{400}{1000}}=1,5(M)$

Bài 5:

$a)\\ CuO+2HCl\to CuCl_2+H_2O\\ b)\\ n_{HCl}=\dfrac{50.3,65\%}{36,5}=0,05(mol)\\ n_{CuO}=\dfrac{1}{2}n_{HCl}=0,025(mol)\\ m_{CuO}=0,025.80=2(g)\\ c)\\ n_{CuCl_2}=n_{CuO}=0,025(mol)\\ m_{dd\ spu}=2+50=52(g)\\ C\%_{dd\ CuCl_2}=\dfrac{0,025.135}{52}.100\%=6,49\%$