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Balance the equation for the reaction between sulfuric acid and sodium hydroxide and include state symbols.

Chemistry

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9 steps

Part d(i)· 5 steps

Balance the equation and include state symbols.

  1. Write the unbalanced equation.

    H2SO4+NaOH→Na2SO4+H2O\text{H}_2\text{SO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}

  2. Balance the sodium (Na) atoms.

    There are 2 Na atoms on the right, so place a coefficient of 2 in front of NaOH.

    H2SO4+2NaOH→Na2SO4+H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}

  3. Balance the hydrogen (H) atoms.

    There are 2(from H2SO4)+2(from 2NaOH)=42 (\text{from H}_2\text{SO}_4) + 2 (\text{from } 2\text{NaOH}) = 4 H atoms on the left. There are 2 H atoms in H2O\text{H}_2\text{O} on the right. Place a coefficient of 2 in front of H2O\text{H}_2\text{O}.

    H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

  4. Check oxygen (O) atoms.

    There are 4(from H2SO4)+2(from 2NaOH)=64 (\text{from H}_2\text{SO}_4) + 2 (\text{from } 2\text{NaOH}) = 6 O atoms on the left. There are 4(from Na2SO4)+2(from 2H2O)=64 (\text{from Na}_2\text{SO}_4) + 2 (\text{from } 2\text{H}_2\text{O}) = 6 O atoms on the right. The equation is balanced.

  5. Add state symbols.

    Sulfuric acid and sodium hydroxide are typically aqueous solutions, sodium sulfate is a soluble salt (aqueous), and water is a liquid.

    H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\boxed{\text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})}

Part d(ii)· 4 steps

Calculate the mass of sodium sulfate produced.

  1. Convert the volume of sulfuric acid from cm3\text{cm}^3 to dm3\text{dm}^3.

    Volume of H2SO4=50 cm3=501000 dm3=0.050 dm3\text{Volume of H}_2\text{SO}_4 = 50 \text{ cm}^3 = \frac{50}{1000} \text{ dm}^3 = 0.050 \text{ dm}^3

  2. Calculate the moles of sulfuric acid (H2SO4\text{H}_2\text{SO}_4).

    Moles of H2SO4=Concentration×Volume\text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} Moles of H2SO4=0.2 mol/dm3×0.050 dm3=0.01 mol\text{Moles of H}_2\text{SO}_4 = 0.2 \text{ mol/dm}^3 \times 0.050 \text{ dm}^3 = 0.01 \text{ mol}

  3. Determine the moles of sodium sulfate (Na2SO4\text{Na}_2\text{SO}_4) produced using the mole ratio from the balanced equation.

    From the balanced equation: H2SO4:Na2SO4\text{H}_2\text{SO}_4 : \text{Na}_2\text{SO}_4 is 1:11:1.

    Moles of Na2SO4=Moles of H2SO4=0.01 mol\text{Moles of Na}_2\text{SO}_4 = \text{Moles of H}_2\text{SO}_4 = 0.01 \text{ mol}

  4. Calculate the mass of sodium sulfate produced.

    Mass of Na2SO4=Moles×RFM\text{Mass of Na}_2\text{SO}_4 = \text{Moles} \times \text{RFM} Mass of Na2SO4=0.01 mol×142 g/mol=1.42 g\text{Mass of Na}_2\text{SO}_4 = 0.01 \text{ mol} \times 142 \text{ g/mol} = 1.42 \text{ g}

    The mass of sodium sulfate produced is 1.42 g\boxed{1.42 \text{ g}}.

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