Question

Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M Na2SO4.

Answer 1

Answer: 43.3 grams

Explanation:-

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}$

Thus $\text{no of moles}of Na_2SO_4={0.0664M}\times {3.06 L}=0.203$

Thus $\text{no of moles}of Ba(OH)_2={0.0820M}\times {2.27L}=0.186$

$Ba(OH)_2(aq)+Na_2SO_4(aq)\rightarrow 2NaOH(aq)+BaSO_4(s)$

As 1 mole of $Ba(OH)_2$ reacts with 1 mole of $Na_2SO_4$

0.186 moles of $Ba(OH)_2$ reacts with =$\frac{1}{1}\times 0.186=0.186moles$  of $Na_2SO_4$

Thus $Ba(OH)_2$ is the limiting reagent and will limit the formation of the products.

As 1 mole of $Ba(OH)_2$ gives with 1 mole of $BaSO_4$ precipitate

0.186 moles of $Ba(OH)_2$ reacts with =$\frac{1}{1}\times 0.186=0.186moles$  of $BaSO_4$ precipitate

Mass of $BaSO_4=moles\times {\text {Molar mass}}=0.186\times 233=43.3g$

Thus mass of the precipitate formed  is 43.3 grams.

Answer 2

We are given with two reactants :barium hydroxide and sodium sulfate. The products of the reaction via double replacement is barium sulfate and sodium hydroxide. according to the solubility rules the product barium sulfate is insoluble. There is a one to one correspondence to every compound. In this case, we just have to find the limiting reactant and base the calculations there. The mass of the precipitate formed is 0.186 grams BaSO4