Bài 13:
nO2= 32/32=1(mol)
a) PTHH: 2 Mg + O2 -to-> 2 MgO
nMg=nMgO=nO2.2=2(mol)
=> mMg= 2.24=48(g)
b) mMgO=40.2=80(g)
Bài 14:
nCaCl2= 55,5/111= 0,5(mol)
a) PTHH: Ca +2 HCl -> CaCl2 + H2
Ta có: nH2=nCa=0,5(mol); nHCl=2.0,5=1(mol)
=> mHCl=1.36,5=36,5(g)
mCa= 40.0,5=20(g)
b) V(H2,đktc)=0,5.22,4=11,2(l)