Answer :
Final answer:
In the chemical reaction of barium chloride with sodium sulfate, about 41.5 grams of barium sulfate can be produced from 37.1 grams of barium chloride according to stoichiometric calculations.
Explanation:
Given the balanced equation BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl, one can find out the amount of barium sulfate produced using the concept of stoichiometry.
First, we need to find out the molar masses of both Barium Chloride (BaCl₂) and Barium Sulfate (BaSO₄).
The molar mass of BaCl₂= 208.23 g/mol
And for BaSO₄, molar mass = 233.43 g/mol
From the balanced chemical equation, we can observe that 1 mole of BaCl₂ reacts to produce 1 mole of BaSO₄.
So, as per the given condition,
If we have 37.1 g of BaCl₂, the moles of BaCl₂ we have would be calculated as follows-
⇒ 37.1 g / 208.23 g/mol = 0.178 moles.
Since the reaction ratio is 1:1, we will also get 0.178 moles of BaSO₄.
Now, to find the mass of BaSO₄ produced, we multiply the moles of BaSO₄ by its molar mass.
⇒ 0.178 moles × 233.43 g/mol = approximately 41.5 g.
Hence, from 37.1 g of BaCl₂, we can produce about approximately 41.5 g of BaSO₄.
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