1.
$n_{MnO_2}=\dfrac{30,45}{87}=0,35(mol)$
$MnO_2+4HCl\to MnCl_2+Cl_2+2H_2O$
$\to n_{Cl_2}=n_{MnO_2}=0,35(mol)$
$\to V=0,35.22,4=7,84l$
2.
$n_{Cl_2}=\dfrac{3,36}{22,4}=0,15(mol)$
$n_{Fe}=\dfrac{6,72}{56}=0,12(mol)$
$2Fe+3Cl_2\buildrel{{t^o}}\over\to 2FeCl_3$
$\to Fe$ dư, $Cl_2$ hết
$n_{FeCl_3}=\dfrac{2}{3}n_{Fe}=0,1(mol)$
$\to m=162,5.0,1=16,25g$
3.
Gọi CTHH muối là $RX_n$
$RX_n+nAgNO_3\to R(NO_3)_n+nAgX$
$n_{RX_n}=\dfrac{24,84}{R+nX}(mol)$
$n_{AgX}=\dfrac{50,76}{108+X}(mol)$
$\to \dfrac{24,84n}{R+nX}=\dfrac{50,76}{108+X}$
$\to 50,76(R+nX)=24,84n(108+X)$
$\Leftrightarrow 50,76R+25,92nX=2682,72n$
Thử lần lượt: $X=35,5(Cl), 80(Br), 127(I)$; $n=1,2,3$
$\to X=80, n=2, R=24(Mg)$
Vậy muối là $MgBr_2$