Question

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

Answer 1

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