what volume of oxygen at R.T.P is required .....?

what volume of oxygen at R.T.P is required a)to burn 5mol of Magnesium b)burn 3.6g of Mg c)produce 3mol Magnesium oxide d)produce 14.4g of magnesium oxide

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  • 10 years ago
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    Magnesium reacts with oxygen according to the reaction equation:

    2 Mg + O → 2 MgO

    That means you need one half mole of oxygen per mole of magnesium burned as well as per mole of magnesium oxide produced.

    The required volume can be found using ideal gas law

    p∙V = n∙R∙T

    =>

    V = n∙R∙T/p

    At room temperature and pressure:

    T = 25°C = 298.15K

    p = 1atm = 101325Pa

    a)

    To burn 5 mole of magnesium you need 2.5 moles of oxygen. So the required volume is:

    V = 2.5mol ∙ 8.314472 Pa∙m³∙mol⁻¹∙K⁻¹ ∙ 298.15K / 101325Pa

    = 0.0612 m³

    = 61.2 L

    b)

    The number of moles of magnesium is:

    n = m/M = 3.6g / 24.305 g∙mol⁻¹ = 0.148 mol

    So you need 0.074 moles of oxygen, which occupy at R.T.P. a volume of:

    V = 0.074mol ∙ 8.314472 Pa∙m³∙mol⁻¹∙K⁻¹ ∙ 298.15K / 101325Pa

    = 1.81×10⁻³ m³

    = 1.81 L

    c)

    To produce 3moles of MgO you need 1.5 moles of oxygen.

    Hence,

    V = 1.5mol ∙ 8.314472 Pa∙m³∙mol⁻¹∙K⁻¹ ∙ 298.15K / 101325Pa

    = 0.0367 m³

    = 36.2 L

    d)

    The number of moles of magnesium oxides produced is:

    n = m/M = 14.4g / 40.3044 g∙mol⁻¹ = 0.3572 mol

    So you need 0.1786 moles of oxygen, which occupy at R.T.P. a volume of:

    V = 0.1786mol ∙ 8.314472 Pa∙m³∙mol⁻¹∙K⁻¹ ∙ 298.15K / 101325Pa

    = 4.37×10⁻³ m³

    = 4.37 L

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  • 3 years ago

    The balanced equation might desire to be: C3H8 + 5 O2 --> 3 CO2 + 4 H2O the important situation to bear in mind is that equivalent volumes of any gases on the comparable temperature and stress comprise the comparable style of moles of molecules. So, in view which you're burning 0.556 L of propane, you may desire to choose 5 X 0.556 = 2.seventy 8 L of O2.

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