Q3.106P - Chemistry: Molecular Nature Of Matter And Change

Question

How many grams of solid barium sulfate form when 35.0 mL of 0.160 M barium chloride reacts with 58.0 mL of 0.065 M sodium sulfate? Aqueous sodium chloride forms also.

Step-by-Step Solution

Answer

When a 35.0 mL of 0.160 M barium chloride reacts with a 58.0 mL of 0.065 M sodium sulfate, 0.88 g of BaSO₄will form.

Step 1: Calculating the number of moles of BaSO 4

Calculate the mole number produced using the given reactants:

$\begin{array}{rcl}Moles of BaS{O}_4& =& 35\times {10}^{-3}L BaC{l}_{2}so\ln \times \frac{0.16 mol BaC{l}_2}{1 L BaC{l}_2 so\ln }\times \frac{1 mol BaS{O}_4}{1 mol BaC{l}_2}\\ & =& 5.6\times {10}^{-3}mol BaS{O}_4\\ & =& 58\times {10}^{-3}L NaS{O}_{4}so\ln \times \frac{0.065 mol NaS{O}_4}{1 L NaS{O}_4 so\ln }\times \frac{1 mol BaS{O}_4}{1 mol NaS{O}_4}\\ & =& 3.77\times {10}^{-3}mol BaS{O}_4.\end{array}$

Step 2: Calculating the mass of BaSO 4

Multiply the molar mass:

$\begin{array}{rcl}Mass of BaS{O}_4& =& 3.77\times {10}^{-3}mol BaS{O}_4\times \frac{233.38 g BaS{O}_4}{1 mol BaS{O}_4}\\ & =& 0.88 g BaS{O}_4.\end{array}$