Consider this reaction 2Mg (s) + O2 (g) - --> 2MgO (s) What volume (in milliners) of gas is required to react with 4.03 g Mg at STP?

A) 1860 mL

B) 2880 mL

C) 3710 mL

D) 45,100 mL

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

by cross multiplication = 0.0895 moles x22.4 l / 1 mole=1.8570 L

into Ml = 1.8570 x1000=1856 ml approximately to 1860