Bài 6:

Tóm tắt:

Cho $m_{Al_2(SO_4)_3}=68,4(g)$

Tính:

$m_{BaCl_2}=?(g)$

$m_{BaSO_4}=?(g)$

$m_{AlCl_3}=?(g)$

Giải:

$n_{Al_2(SO_4)_3}=\dfrac{68,4}{342}=0,2(mol)$

$PTHH:Al_2(SO_4)_3+3BaCl_2\to 2AlCl_3+3BaSO_4\downarrow$

Theo PT: $n_{BaCl_2}=n_{BaSO_4}=0,6(mol);n_{AlCl_3}=0,4(mol)$

$\Rightarrow m_{BaCl_2}=0,6.208=124,8(g)$

$m_{BaSO_4}=0,6.233=139,8(g)$

$m_{AlCl_3}=0,4.133,5=53,4(g)$

Bài 7:

Tóm tắt:

Cho $m_{AgNO_3}=51(g)$

Tính:

$m_{AlCl_3}=?(g)$

$m_{Al(NO_3)_3}=?(g)$

$m_{AgCl}=?(g)$

Giải:

$n_{AgNO_3}=\dfrac{51}{170}=0,3(mol)$

$PTHH:3AgNO_3+AlCl_3\to 3AgCl\downarrow+Al(NO_3)_3$

Theo PT: $n_{AlCl_3}=n_{Al(NO_3)_3}=0,1(mol);n_{AgCl}=0,3(mol)$

$\Rightarrow m_{AlCl_3}=0,1.133,5=13,35(g)$

$m_{Al(NO_3)_3}=0,1.213=21,3(g)$

$m_{AgCl}=0,3.143,5=43,05(g)$