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Calculate the mass of sodium oxide, Na2O, formed when 3.45 g of sodium burns in air. (O = 16, Na = 23)

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4.65 g\text{4.65 g}

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I will solve question 31.

31. Calculate the mass of sodium oxide, Na2O\text{Na}_2\text{O}, formed when 3.45 g of sodium burns in air. (O=16\text{O} = 16, Na=23\text{Na} = 23)

  1. Write the balanced chemical equation for the reaction.

    Sodium reacts with oxygen in the air to form sodium oxide.

    4Na(s)+O2(g)⟶2Na2O(s)4\text{Na(s)} + \text{O}_2\text{(g)} \longrightarrow 2\text{Na}_2\text{O(s)}

  2. Calculate the molar mass of sodium (Na\text{Na}) and sodium oxide (Na2O\text{Na}_2\text{O}).

    Molar mass of Na=23 g/mol\text{Na} = 23 \text{ g/mol}
    Molar mass of Na2O=(2×23)+(1×16)=46+16=62 g/mol\text{Na}_2\text{O} = (2 \times 23) + (1 \times 16) = 46 + 16 = 62 \text{ g/mol}

  3. Calculate the moles of sodium given.

    Moles of Na=Mass of NaMolar mass of Na=3.45 g23 g/mol=0.15 mol\text{Moles of Na} = \frac{\text{Mass of Na}}{\text{Molar mass of Na}} = \frac{3.45 \text{ g}}{23 \text{ g/mol}} = 0.15 \text{ mol}

  4. Use the mole ratio from the balanced equation to find the moles of sodium oxide formed.

    From the balanced equation, 4 moles of Na\text{Na} produce 2 moles of Na2O\text{Na}_2\text{O}.
    So, the mole ratio of Na:Na2O\text{Na} : \text{Na}_2\text{O} is 4:24:2 or 2:12:1.

    Moles of Na2O=Moles of Na×2 mol Na2O4 mol Na=0.15 mol×12=0.075 mol\text{Moles of Na}_2\text{O} = \text{Moles of Na} \times \frac{2 \text{ mol Na}_2\text{O}}{4 \text{ mol Na}} = 0.15 \text{ mol} \times \frac{1}{2} = 0.075 \text{ mol}

  5. Calculate the mass of sodium oxide formed.

    Mass of Na2O=Moles of Na2O×Molar mass of Na2O\text{Mass of Na}_2\text{O} = \text{Moles of Na}_2\text{O} \times \text{Molar mass of Na}_2\text{O} Mass of Na2O=0.075 mol×62 g/mol=4.65 g\text{Mass of Na}_2\text{O} = 0.075 \text{ mol} \times 62 \text{ g/mol} = 4.65 \text{ g}

    The mass of sodium oxide formed is 4.65 g\text{4.65 g}.

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