Barium Chloride & Sodium Sulfate: Calculating BaSO? Yield

Step 1 of 2
Here, we have to calculate the grams of Barium sulfate.
First, let’s write the balanced chemical equation.
\(\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{BaSO}_{4}(\mathrm{~s})+2 \mathrm{NaCl}(\mathrm{s})\)

Let’s calculate the reacted mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)
Molar mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)
Atomic mass of \(\mathrm{Na}=2 \times \mathrm{Na}=2 \times 22.9 \mathrm{~g} / \mathrm{mol}=45.8 \mathrm{~g} / \mathrm{mol}\)
Atomic mass of \(\mathrm{S}=1 \times \mathrm{S}=1 \times 32.0 \mathrm{~g} / \mathrm{mol}=32.0 \mathrm{~g} / \mathrm{mol}\)
Atomic mass of \(O=4 \times O=4 \times 16 \mathrm{~g} / \mathrm{mol}=64 \mathrm{~g} / \mathrm{mol}\)
                                                                            
Molar mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)                           
                                                                   
\(141.8 \equiv 142 \mathrm{~g} / \mathrm{mol}\)
                                                                            
\(\text { Mass of } \mathrm{Na}_{2} \mathrm{SO}_{4}=\frac{\text { Molarmass of } \mathrm{Na}_{2} \mathrm{SO}_{4} \times \text { Volume } \times \text { Molarity }}{1000}\)
Given:
Volume of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)  = 58 ml
Molarity of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) = 0.065M.
                    
\( =\frac{142 \times 58 \times 0.065 M}{1000}=0.53534 \mathrm{gm}\) .
Therefore, Mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is 0.53534 gm.