Q3.107P - Chemistry: Molecular Nature Of Matter And Change

Question

How many moles of excess reactant are present when 350 mL of 0.210 M sulfuric acid reacts with 0.500 L of 0.196 M sodium hydroxide to form water and aqueous sodium sulfate?

Step-by-Step Solution

Answer

A 2.45×10^-2mol H₂SO₄ of excess reactant are present when 350 mL of 0.210 M sulfuric acid reacts with 0.500 L of 0.196 M sodium hydroxide to form water and aqueous sodium sulfate.

Step 1: Calculating the number of moles of BaSO 4

 Calculate the mole number produced using the given reactants:

$\begin{array}{rcl}Moles of N{a}_{2}S{O}_4& =& 35\times {10}^{-3}L H_{2}S{O}_{4}so\ln \times \frac{0.21 mol H_{2}S{O}_4}{1 L H_{2}S{O}_{4}so\ln }\times \frac{1 mol N{a}_{2}S{O}_4}{1 mol H_{2}S{O}_4}\\ & =& 0.0735 mol N{a}_{2}S{O}_4\backslash \mathrm{hfill}\\ & =& 50.5 l NaOHso\ln \times \frac{0.196 mol NaOH}{1 L NaOH so\ln }\times \frac{1 mol N{a}_{2}S{O}_4}{2 mol NaOH}\\ & =& 0.049mol N{a}_{2}S{O}_4.\end{array}$

Step 2: Calculating the mass of BaSO 4

Substract the above result from the initial number of moles:

$\begin{array}{rcl}{H}_{2}S{O}_4 excess& =& 35\times {10}^{-3}L H_{2}S{O}_4 so\ln \times \frac{0.21 mol H_{2}S{O}_4}{1 L H_{2}S{O}_4 so\ln }-0.049 mol H_{2}S{O}_4\\ & =& 2.45\times {10}^{-2} mol.\\ & & \end{array}$