The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
the moles 02=0.1679 x1/20.0829 moles
0.0895 moles=? L
into Ml = 1.8570 x1000=1856 ml approximately to 1860