Question

$H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$

In order to produce 28.4 g of $Na_2SO_4$ how many grams of $H_2SO_4$ and $NaOH$ must be taken ?

Answer 1

19.6 g of $H_2SO_4$ and 16.0 g of $NaOH$

Answer 2

1. Calculate Molar Masses:

   • $Na_2SO_4$: (2 $\times$ 23) + 32 + (4 $\times$ 16) = 142 g/mol
   • $H_2SO_4$: (2 $\times$ 1) + 32 + (4 $\times$ 16) = 98 g/mol
   • $NaOH$: 23 + 16 + 1 = 40 g/mol

2. Calculate Moles of $Na_2SO_4$: $n(Na_2SO_4) = \frac{28.4 \text{ g}}{142 \text{ g/mol}} = 0.2 \text{ mol}$

3. Determine Moles of Reactants using Stoichiometry: The balanced equation is $H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$.

   • From the equation, 1 mole of $H_2SO_4$ is required for 1 mole of $Na_2SO_4$. Moles of $H_2SO_4$ required = 0.2 mol.
   • From the equation, 2 moles of $NaOH$ are required for 1 mole of $Na_2SO_4$. Moles of $NaOH$ required = 2 $\times$ 0.2 mol = 0.4 mol.

4. Calculate Mass of Reactants:

   • Mass of $H_2SO_4$ = moles $\times$ molar mass = 0.2 mol $\times$ 98 g/mol = 19.6 g
   • Mass of $NaOH$ = moles $\times$ molar mass = 0.4 mol $\times$ 40 g/mol = 16.0 g