Câu 15:
$n_{Cr}=\dfrac{7,8}{52}=0,15(mol)$
$2Cr+3Cl_2\buildrel{{t^o}}\over\to 2CrCl_3$
$\Rightarrow n_{Cl_2}=\dfrac{3}{2}n_{Cr}=0,225(mol)$
$V_{Cl_2}=0,225.22,4=5,04l$
$\to$ chọn $C$
Câu 16:
Bảo toàn khối lượng:
$m_{Cl_2}=40,3-11,9=28,4g$
$\Rightarrow n_{Cl_2}=\dfrac{28,4}{71}=0,4(mol)$
$\Rightarrow V_{Cl_2}=0,4.22,4=8,96l$
$\to$ chọn $A$
