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A) State and explain the observations made when burning magnesium is lowered into a glass jar of carbon (IV) oxide. b) Write an equation for the reaction that takes place in (a). 22. Calculate the mol

Chemistry

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b
90.1

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21.

Part a

State and explain the observations made when burning magnesium is lowered into a glass jar of carbon (IV) oxide.

  • Observations: The magnesium continues to burn with a bright white flame. A white solid (magnesium oxide) and black specks (carbon) are formed on the inside of the jar.
  • Explanation: Magnesium is a highly reactive metal that is able to reduce carbon (IV) oxide (CO2\text{CO}_2) to elemental carbon, while itself being oxidized to magnesium oxide. The reaction is highly exothermic, allowing the magnesium to continue burning despite the absence of oxygen.

Part b90.1

Write an equation for the reaction that takes place in (a)

2Mg(s)+CO2(g)→2MgO(s)+C(s)2\text{Mg}(s) + \text{CO}_2(g) \rightarrow 2\text{MgO}(s) + \text{C}(s)

22.
(Assuming the atomic mass of Hydrogen (H) is 1.0, as H=10. is a likely typo.)

  1. Calculate the molar mass of sodium hydroxide (NaOH).

    Mr(NaOH)=Na+O+H=23.0+16.0+1.0=40.0 g/mol\text{M}_r(\text{NaOH}) = \text{Na} + \text{O} + \text{H} = 23.0 + 16.0 + 1.0 = 40.0 \text{ g/mol}

  2. Calculate the molarity of the sodium hydroxide solution.

    The concentration is 4.0 g/L4.0 \text{ g/L}.

    Molarity of NaOH=Mass concentrationMolar mass=4.0 g/L40.0 g/mol=0.1 mol/L\text{Molarity of NaOH} = \frac{\text{Mass concentration}}{\text{Molar mass}} = \frac{4.0 \text{ g/L}}{40.0 \text{ g/mol}} = 0.1 \text{ mol/L}

  3. Calculate the moles of sodium hydroxide used.

    Volume of NaOH solution =22.2 cm3=0.0222 L= 22.2 \text{ cm}^3 = 0.0222 \text{ L}.

    Moles of NaOH=Molarity×Volume=0.1 mol/L×0.0222 L=0.00222 mol\text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.1 \text{ mol/L} \times 0.0222 \text{ L} = 0.00222 \text{ mol}

  4. Determine the moles of the dibasic acid.

    A dibasic acid (H2A\text{H}_2\text{A}) reacts with sodium hydroxide in a 1:2 mole ratio:

    H2A(aq)+2NaOH(aq)→Na2A(aq)+2H2O(l)\text{H}_2\text{A}(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{A}(aq) + 2\text{H}_2\text{O}(l) Moles of acid=Moles of NaOH2=0.00222 mol2=0.00111 mol\text{Moles of acid} = \frac{\text{Moles of NaOH}}{2} = \frac{0.00222 \text{ mol}}{2} = 0.00111 \text{ mol}

  5. Calculate the relative formula mass of the dibasic acid.

    Mass of dibasic acid =0.1 g= 0.1 \text{ g}.

    Relative formula mass=Mass of acidMoles of acid=0.1 g0.00111 mol≈90.09\text{Relative formula mass} = \frac{\text{Mass of acid}}{\text{Moles of acid}} = \frac{0.1 \text{ g}}{0.00111 \text{ mol}} \approx 90.09

    Rounding to three significant figures:

    Relative formula mass=90.1\text{Relative formula mass} = \boxed{90.1}
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