volume calculation (1 Viewer)

[Halo189]

1.20g of magnesium burns in air to form magnesium oxide
2Mg+O2--->2MgO

If oxygen makes up 20% of the air, calculate the volume of air required for the complete combustion of the 1.20g of magnesium

 

[InteGrand]

1.20g of magnesium burns in air to form magnesium oxide
2Mg+O2--->2MgO

If oxygen makes up 20% of the air, calculate the volume of air required for the complete combustion of the 1.20g of magnesium

(HSC Chemistry data page: http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/chemistry-data-sheet.pdf)

1.20 g of magnesium is n molecules of magnesium, where n = m/M = 1.20 g/(24.3050 g/mol) ≈ 0.0493725571 mol.

So the number of molecules of O₂ (oxygen gas) required is half of this (as seen from the stoichiometry), i.e. about 0.0246862785 mol.

According to the HSC Chemistry data page, 1 mol of idealised gas at 100 kPa and 25°C has a volume of 24.79 L.

So 0.0246862785 mol in these conditions has a volume of ~ 0.0246862785 mol × 24.79 L/mol ≈ 611.972844 mL.

So we need 611.972844 mL of O₂ gas. Since oxygen is one-fifth the volume of air (given), the required volume of air is approximately 5×611.972844 mL = 3.05986422 L, or 3 L (to 1 sig. fig.).

 

[Halo189]

(HSC Chemistry data page: http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/chemistry-data-sheet.pdf)

1.20 g of magnesium is n molecules of magnesium, where n = m/M = 1.20 g/(24.3050 g/mol) ≈ 0.0493725571 mol.

So the number of molecules of O₂ (oxygen gas) required is half of this (as seen from the stoichiometry), i.e. about 0.0246862785 mol.

According to the HSC Chemistry data page, 1 mol of idealised gas at 100 kPa and 25°C has a volume of 24.79 L.

So 0.0246862785 mol in these conditions has a volume of ~ 0.0246862785 mol × 24.79 L/mol ≈ 611.972844 mL.

So we need 611.972844 mL of O₂ gas. Since oxygen is one-fifth the volume of air (given), the required volume of air required is aproximately 5×611.972844 mL = 3.05986422 L, or 3 L (to 1 sig. fig.).

Thanks heaps