$n_{C_2H_4}=\dfrac{5,6}{28}=0,2(mol)$
$C_2H_4+3O_2\xrightarrow{{t^o}} 2CO_2+2H_2O$
$\to n_{O_2}=3n_{C_2H_4}=0,2.3=0,6(mol)$
$\to V_{O_2}=0,6.22,4=13,44l$
$n_{CO_2}=2n_{C_2H_4}=0,4(mol)$
$CO_2+Ca(OH)_2\to CaCO_3\downarrow+H_2O$
$\to n_{CaCO_3\downarrow}=n_{CO_2}=0,4(mol)$
$\to m_{\downarrow}=0,4.100=40g$