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djyliett [7]
2 years ago
8

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.

Chemistry
2 answers:
Trava [24]2 years ago
7 0

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.

kicyunya [14]2 years ago
6 0
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 
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Answer:

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Explanation:

Bismuth(iii) sulfate is an ionic compound therefore, their is transfer of electron. Ionic compound has both cations and anions. The cations is positively charged ion while the anions is negatively charged ions. The cations loses electron to become positively charged while the anions gains electron to become negatively charged.

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