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The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
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