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k0ka [10]
3 years ago
15

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
Chemistry
1 answer:
nydimaria [60]3 years ago
6 0

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

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  • Particles could refer to atoms, molecules, formula units.
  • <em>Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).</em>

<em><u>Using cross multiplication:  </u></em>

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??? mole → 8.95 x 10²³ molecules.

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