a,
Hiện tượng: có kết tủa trắng xuất hiện.
$CaSO_4+BaCl_2\to BaSO_4+CaCl_2$
b,
$n_{CaSO_4}=\dfrac{1,11}{136}=0,0082(mol)$
$n_{BaCl_2}=\dfrac{2,33}{208}=0,011(mol)$
$\Rightarrow BaCl_2$ dư
$n_{BaSO_4}=n_{CaSO_4}=0,0082(mol)$
$\Rightarrow m_{\downarrow}=233.0,0082=1,9106g$
c,
$V_{dd}=100ml=0,1l$
$n_{CaCl_2}=n_{CaSO_4}=0,0082(mol)$
$\to C_{M_{CaCl_2}}=\dfrac{0,0082}{0,1}=0,082M$
$n_{BaCl_2}=0,011-0,0082=0,0028(mol)$
$\to C_{M_{BaCl_2}}=\dfrac{0,0028}{0,1}=0,028M$